26.03.2020

# how many moles are in 150 grams of C6H12O6 4

### Faq

Chemistry

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of over Nine moles of Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of .

Answer 4 : Mass of = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

Solution

Solution 1 : Given,

Given mass of = 55 g

Molar mass of = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of = = = 0.344 moles

Balanced chemical reaction is, From the given reaction, we conclude that

1 mole of gives              →         3 moles of CO

0.344 moles of gives    →         3 × 0.344 moles of CO

=         1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

= 1.032 × 28.01

= 28.90 g

Solution 2 : The balanced chemical reaction is, From the given reaction, we conclude that the Six moles of over Nine moles of is the correct option.

Solution 3 : The balanced chemical reaction is, From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of .

Solution 4 : Given,

Given mass of = 150 g

Molar mass of = 232.29 g/mole

Molar mass of = 31.998 g/mole

Moles of = = The balanced chemical equation is, From the given balanced equation, we conclude that

1 mole of gives          →       3 moles of  of gives  → of Mass of = Number of moles of × Molar mass of = Therefore, the mass of = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of = 4.2 moles

Balanced chemical equation is, From the given chemical reaction, we conclude that

1 mole of obtained from 2 moles of NaCN

4.2 moles of obtained   →   2 × 4.2 moles of NaCN

Therefore,

The moles of NaCN needed = 2 × 4.2 = 8.4 moles

Chemistry

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of over Nine moles of Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of .

Answer 4 : Mass of = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Answer 5 : 8.4 moles of sodium cyanide (NaCN) would be needed.

Solution

Solution 1 : Given,

Given mass of = 55 g

Molar mass of = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of = = = 0.344 moles

Balanced chemical reaction is, From the given reaction, we conclude that

1 mole of gives              →         3 moles of CO

0.344 moles of gives    →         3 × 0.344 moles of CO

=         1.032 moles

Mass of CO = Number of moles of CO × Molar mass of CO

= 1.032 × 28.01

= 28.90 g

Solution 2 : The balanced chemical reaction is, From the given reaction, we conclude that the Six moles of over Nine moles of is the correct option.

Solution 3 : The balanced chemical reaction is, From the given balanced reaction, we conclude that Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of .

Solution 4 : Given,

Given mass of = 150 g

Molar mass of = 232.29 g/mole

Molar mass of = 31.998 g/mole

Moles of = = The balanced chemical equation is, From the given balanced equation, we conclude that

1 mole of gives          →       3 moles of  of gives  → of Mass of = Number of moles of × Molar mass of = Therefore, the mass of = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of = 4.2 moles

Balanced chemical equation is, From the given chemical reaction, we conclude that

1 mole of obtained from 2 moles of NaCN

4.2 moles of obtained   →   2 × 4.2 moles of NaCN

Therefore,

The moles of NaCN needed = 2 × 4.2 = 8.4 moles

Mathematics
8)2.258×10^24 molecules
Mathematics
8)2.258×10^24 molecules
Chemistry

1. 10 moles of NO

2. 25 moles of NaCl

3. 1200 moles CO₂

4. 1.042 mole of MgO

5. 0.714 moles H₂ gas

6. 1040 g of BaCl₂

7. 9.5 g

8. 45.44 g of Au

9. 15 g of AlCl₃

Explanation:

Ans 1.

Data Given:

moles of Oxygen = 5 moles

moles of nitrogen monoxide = ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

N₂   +   O₂  -----------> 2NO

1 mol                 2 mol

So if we look at the reaction  1 mole Oxygen (O₂)  gives 2 moles of nitrogen monoxide NO, then how many moles of nitrogen monoxide will be produced by 5 moles of Oxygen (O₂)

For this apply unity formula

1 mole of O₂ ≅ 2 moles of NO

5 mole of O₂ ≅ X moles of NO

By Doing cross multiplication

moles of NO = 2 moles x 5 moles / 1 mole

moles of NO = 10 mole

5 mole of O₂ will produce 10 moles of NO

________________

Ans 2.

Data Given:

moles of HCl = 25 moles

moles of NaCl = ?

Solution:

To solve this problem we have to look at the reaction

Neutralization Reaction:

HCl   +   NaOH  -----------> NaCl + H₂O

1 mol                                     1 mol

So if we look at the reaction  1 mole  HCl  gives 1 moles of NaCl, then how many moles of NaCl will be formed by 25 moles of HCl

For this apply unity formula

1 mole of HCl ≅ 1 moles of NaCl

25 mole of HCl ≅ X moles of NaCl

By Doing cross multiplication

moles of NaCl = 1 moles x 25 moles / 1 mole

moles of NaCl = 25 mole

25 mole of HCL will form 25 moles of NaCl

________________

Ans 3.

Data Given:

moles of C₈H₁₈ = 150 moles

moles of CO₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2C₈H₁₈   +  25O₂  -----------> 16CO₂ + 18H₂O

2 mol                                     16 mol

So if we look at the reaction  2 mole C₈H₁₈ gives 16 moles of CO₂, then how many moles of CO₂ will be Produce by 150 moles of C₈H₁₈

For this apply unity formula

2 mole of C₈H₁₈ ≅ 16 moles of CO₂

150 mole of C₈H₁₈ ≅ X moles of CO₂

By Doing cross multiplication

moles of CO₂= 16 moles x 150 moles / 2 mole

moles of CO₂ =  1200 mole

150 mole of C₈H₁₈ will form 1200 moles of CO₂

______________________

Ans 4.

Data Given:

mass of Mg = 25 g

moles of MgO= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2Mg + O₂  -----------> 2MgO

2 mol                        2 mol

Convert moles of Mg to mass

Molar mass of Mg = 24 g/mol

So,

2Mg        +       O₂  ----------->     2MgO

2 mol (24 g/mol)                             2 mol

48 g                                          2 mol

So if we look at the reaction 48 g of Mg gives 2 moles of MgO, then how many moles of MgO will be Produce by 25 g of Mg

For this apply unity formula

48 g of Mg ≅ 2 moles of MgO

25 g of Mg ≅ X moles of MgO

By Doing cross multiplication

moles of MgO =  2 moles x 25 g / 48 g

moles of MgO = 1.042 mole

25 g of Mg will form 1.042 moles of MgO

______________________

Ans 5.

Data Given:

mass of Li = 10 g

moles of H₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2Li + 2 H₂O  -----------> 2LiOH  +  H₂

2 mol                                           1 mol

Convert moles of Li to mass

Molar mass of Li = 7 g/mol

So,

2Li      +       2H₂O  -----------> 2LiOH  +  H₂

2 mol (7 g/mol)                                            1 mol

14 g                                                          1 mol

So if we look at the reaction 14 g of Li gives 1 moles of H₂ gas, then how many moles of H₂ gas will be Produce by 10 g of Li

For this apply unity formula

14 g of Li ≅ 1 moles of H₂

10 g of Li ≅ X moles of H₂

By Doing cross multiplication

moles of H₂ =  1 moles x 10 g / 14 g

moles of H₂ = 0.714 mole

10 g of Li will form 0.714 moles of H₂

______________________

Ans 6.

Data Given:

moles of Na₂SO₄= 5 moles

mass of BaCl₂ = ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

Na₂SO₄ + BaCl₂  -----------> 2NaCl + BaSO₄

1 mol            1 mol

Convert moles of BaCl₂ to mass

Molar mass of BaCl₂ = 208 g/mol

So,

Na₂SO₄    +        BaCl₂  -----------> 2NaCl + BaSO₄

1 mol            1 mol (208 g/mol)

1 mol                   208 g

So if we look at the reaction 1 mole of Na₂SO₄ react with 208 g of BaCl₂, Then how many grams of  BaCl₂ will react with 5 moles of Na₂SO₄

For this apply unity formula

1 mole of Na₂SO₄ ≅ 208 g of BaCl₂

5 mole of Na₂SO₄≅ X g of BaCl₂

By Doing cross multiplication

mass of BaCl₂ =  208 g x  5 moles / 1 mole

mass of BaCl₂ = 1040 g

5 moles of Na₂SO₄ will react with 1040 g of BaCl₂

______________________

Ans 7.

Data Given:

mass of MgCO₃ = 20 g

moles of MgO= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

MgCO₃ -----------> MgO + CO₂

1 mol                     1 mol

Convert moles to mass

Molar mass of MgCO₃ = 84 g/mol

Molar mass of MgO = 40 g/mol

So,

MgCO₃      ------------->    MgO        +       CO₂

1 mol (84 g/mol)                 1 mol (40 g/mol)

84 g                                     40 g

So if we look at the reaction 84 g of MgCO₃ gives 40 g of MgO, then how many g of MgO will be Produce by 20 g of MgCO₃

For this apply unity formula

84 g of MgCO₃ ≅ 40 g of MgO

20 g of MgCO₃ ≅ X g of MgO

By Doing cross multiplication

mass of MgO = 40 g x 20 g / 84g

mass of MgO = 9.5 g

20 g of MgCO₃ will produce 9.5 g of MgO

________________

The remaining portion is in attachment  Chemistry

1. 10 moles of NO

2. 25 moles of NaCl

3. 1200 moles CO₂

4. 1.042 mole of MgO

5. 0.714 moles H₂ gas

6. 1040 g of BaCl₂

7. 9.5 g

8. 45.44 g of Au

9. 15 g of AlCl₃

Explanation:

Ans 1.

Data Given:

moles of Oxygen = 5 moles

moles of nitrogen monoxide = ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

N₂   +   O₂  -----------> 2NO

1 mol                 2 mol

So if we look at the reaction  1 mole Oxygen (O₂)  gives 2 moles of nitrogen monoxide NO, then how many moles of nitrogen monoxide will be produced by 5 moles of Oxygen (O₂)

For this apply unity formula

1 mole of O₂ ≅ 2 moles of NO

5 mole of O₂ ≅ X moles of NO

By Doing cross multiplication

moles of NO = 2 moles x 5 moles / 1 mole

moles of NO = 10 mole

5 mole of O₂ will produce 10 moles of NO

________________

Ans 2.

Data Given:

moles of HCl = 25 moles

moles of NaCl = ?

Solution:

To solve this problem we have to look at the reaction

Neutralization Reaction:

HCl   +   NaOH  -----------> NaCl + H₂O

1 mol                                     1 mol

So if we look at the reaction  1 mole  HCl  gives 1 moles of NaCl, then how many moles of NaCl will be formed by 25 moles of HCl

For this apply unity formula

1 mole of HCl ≅ 1 moles of NaCl

25 mole of HCl ≅ X moles of NaCl

By Doing cross multiplication

moles of NaCl = 1 moles x 25 moles / 1 mole

moles of NaCl = 25 mole

25 mole of HCL will form 25 moles of NaCl

________________

Ans 3.

Data Given:

moles of C₈H₁₈ = 150 moles

moles of CO₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2C₈H₁₈   +  25O₂  -----------> 16CO₂ + 18H₂O

2 mol                                     16 mol

So if we look at the reaction  2 mole C₈H₁₈ gives 16 moles of CO₂, then how many moles of CO₂ will be Produce by 150 moles of C₈H₁₈

For this apply unity formula

2 mole of C₈H₁₈ ≅ 16 moles of CO₂

150 mole of C₈H₁₈ ≅ X moles of CO₂

By Doing cross multiplication

moles of CO₂= 16 moles x 150 moles / 2 mole

moles of CO₂ =  1200 mole

150 mole of C₈H₁₈ will form 1200 moles of CO₂

______________________

Ans 4.

Data Given:

mass of Mg = 25 g

moles of MgO= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2Mg + O₂  -----------> 2MgO

2 mol                        2 mol

Convert moles of Mg to mass

Molar mass of Mg = 24 g/mol

So,

2Mg        +       O₂  ----------->     2MgO

2 mol (24 g/mol)                             2 mol

48 g                                          2 mol

So if we look at the reaction 48 g of Mg gives 2 moles of MgO, then how many moles of MgO will be Produce by 25 g of Mg

For this apply unity formula

48 g of Mg ≅ 2 moles of MgO

25 g of Mg ≅ X moles of MgO

By Doing cross multiplication

moles of MgO =  2 moles x 25 g / 48 g

moles of MgO = 1.042 mole

25 g of Mg will form 1.042 moles of MgO

______________________

Ans 5.

Data Given:

mass of Li = 10 g

moles of H₂= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

2Li + 2 H₂O  -----------> 2LiOH  +  H₂

2 mol                                           1 mol

Convert moles of Li to mass

Molar mass of Li = 7 g/mol

So,

2Li      +       2H₂O  -----------> 2LiOH  +  H₂

2 mol (7 g/mol)                                            1 mol

14 g                                                          1 mol

So if we look at the reaction 14 g of Li gives 1 moles of H₂ gas, then how many moles of H₂ gas will be Produce by 10 g of Li

For this apply unity formula

14 g of Li ≅ 1 moles of H₂

10 g of Li ≅ X moles of H₂

By Doing cross multiplication

moles of H₂ =  1 moles x 10 g / 14 g

moles of H₂ = 0.714 mole

10 g of Li will form 0.714 moles of H₂

______________________

Ans 6.

Data Given:

moles of Na₂SO₄= 5 moles

mass of BaCl₂ = ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

Na₂SO₄ + BaCl₂  -----------> 2NaCl + BaSO₄

1 mol            1 mol

Convert moles of BaCl₂ to mass

Molar mass of BaCl₂ = 208 g/mol

So,

Na₂SO₄    +        BaCl₂  -----------> 2NaCl + BaSO₄

1 mol            1 mol (208 g/mol)

1 mol                   208 g

So if we look at the reaction 1 mole of Na₂SO₄ react with 208 g of BaCl₂, Then how many grams of  BaCl₂ will react with 5 moles of Na₂SO₄

For this apply unity formula

1 mole of Na₂SO₄ ≅ 208 g of BaCl₂

5 mole of Na₂SO₄≅ X g of BaCl₂

By Doing cross multiplication

mass of BaCl₂ =  208 g x  5 moles / 1 mole

mass of BaCl₂ = 1040 g

5 moles of Na₂SO₄ will react with 1040 g of BaCl₂

______________________

Ans 7.

Data Given:

mass of MgCO₃ = 20 g

moles of MgO= ?

Solution:

To solve this problem we have to look at the reaction

Reaction:

MgCO₃ -----------> MgO + CO₂

1 mol                     1 mol

Convert moles to mass

Molar mass of MgCO₃ = 84 g/mol

Molar mass of MgO = 40 g/mol

So,

MgCO₃      ------------->    MgO        +       CO₂

1 mol (84 g/mol)                 1 mol (40 g/mol)

84 g                                     40 g

So if we look at the reaction 84 g of MgCO₃ gives 40 g of MgO, then how many g of MgO will be Produce by 20 g of MgCO₃

For this apply unity formula

84 g of MgCO₃ ≅ 40 g of MgO

20 g of MgCO₃ ≅ X g of MgO

By Doing cross multiplication

mass of MgO = 40 g x 20 g / 84g

mass of MgO = 9.5 g

20 g of MgCO₃ will produce 9.5 g of MgO

________________

The remaining portion is in attachment  Chemistry
1) A
"arrow" should be "yields."
The coefficients are mole ratios. So every 4 moles of NH3, or ammonia, produce 6 moles of H2O, water.

(12 mol NH3)(6 mol H2O/4 mol NH3) = 18 mol H2O produced

2) C
Again, use the coefficients to form mole ratios and solve. See #1.

3) C
4)C
5) C
Moles, clearly. It's another constant you MUST MEMORIZE, like Avogadro's number. (Shame on you if you don't know what Avogadro's number means!) At STP (273.15 K and 1 atm), one mole of any ideal gas is 22.4 L.

6) C
Same thing as #1-4, except with a twist: molar masses. You calculate the molar masses using the periodic table.

Molar mass of H = 1.008 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H2O = 2(1.008 g/mol) + 16.00 g/mol) = 18.016 g/mol
Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol

Divide the mass of water by the molar mass to find the number of moles of water. Use mole ratios to find the number of moles of O2 consumed. Then, multiply by the molar mass to find mass of O2. Tada!

7)
8)
9)
10)
11)
12) See number 13
13) True
14) False; doesn't tell a thing about the speed of the reaction, how spontaneous it is, or at what temperature or pressure it must take place.
Chemistry
1) A
"arrow" should be "yields."
The coefficients are mole ratios. So every 4 moles of NH3, or ammonia, produce 6 moles of H2O, water.

(12 mol NH3)(6 mol H2O/4 mol NH3) = 18 mol H2O produced

2) C
Again, use the coefficients to form mole ratios and solve. See #1.

3) C
4)C
5) C
Moles, clearly. It's another constant you MUST MEMORIZE, like Avogadro's number. (Shame on you if you don't know what Avogadro's number means!) At STP (273.15 K and 1 atm), one mole of any ideal gas is 22.4 L.

6) C
Same thing as #1-4, except with a twist: molar masses. You calculate the molar masses using the periodic table.

Molar mass of H = 1.008 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H2O = 2(1.008 g/mol) + 16.00 g/mol) = 18.016 g/mol
Molar mass of O2 = 2(16.00 g/mol) = 32.00 g/mol

Divide the mass of water by the molar mass to find the number of moles of water. Use mole ratios to find the number of moles of O2 consumed. Then, multiply by the molar mass to find mass of O2. Tada!

7)
8)
9)
10)
11)
12) See number 13
13) True
14) False; doesn't tell a thing about the speed of the reaction, how spontaneous it is, or at what temperature or pressure it must take place.
Chemistry

1) 1.964 moles of O2

2) 0.670 moles of H2SO4

3)  2284.8 liters of oxygen

4) 24.3 grams of S

5) 1864.6 L of hydrogen gas

6) 19.54 grams of HCl

7) 45.0 L of CO2 produced

8) 36.0 L of NO

9) 30L of CO2

Explanation:

1. If 22 L of methane (CH4) gas burns (combustion), how many moles of oxygen will  be needed for complete combustion?

Step 1: The balanced equation:

CH4 + 2O2 → CO2 + 2H2O

Step 2: Calculate moles of CH4

Since 22.4 L = 1.0 mol

22.4L = 0.982 moles of CH4

Step 3: Calculate moles of O2 needed

For 1 mol CH4 we need 2 moles of O2 to produce 1 mol of CO2 and 2 moles of H2O

For 0.982 moles of CH4 we need 2*0.982 = 1.964 moles of O2

2. Acid rain is produced when sulfur trioxide reacts with water in a composition  reaction. If 15 L of sulfur trioxide is present in the air how many moles of sulfuric  acid are produced?

Step 1: The balanced equation

SO3(g) + H2O(l) → H2SO4(aq)

Step 2: Calculate moles of SO3

22.4 L = 1 mol

15.0 L = 0.670 moles

Step 3: Calculate moles of H2SO4

For 1 mol of SO3 we need 1 mol of H2O to produce 1 mol of H2SO4

For 0.670 moles of SO3 we will produce 0.670 moles of H2SO4

3. If calcium chlorate is heated it decomposes into calcium chloride and oxygen gas.

How many liters of oxygen can be produced form heating 34 moles of calcium  carbonate?

Step 1: The balanced equation

Ca(ClO3)2  → CaCl2 + 3O2

Step 2: Calculate moles of O2

For 1 mol of Ca(ClO3)2 consumed, we produce 1 mol of CaCl2 and 3 moles of O2

For 34 moles of Ca(ClO3)2 consumed, we'll have 3*34 = 102 moles of O2

Step 3: Calculate volume of oxygen

1 mol = 22.4 L

102 moles = 22.4 * 102 = 2284.8 liters of oxygen

4. Sulfur and oxygen combine to form sulfur dioxide in a composition reaction. How  many grams of sulfur are required to react with 17 L of oxygen?

Step 1: The balanced equation

S + O2 → SO2

Step 2: Calculate number of moles of O2

22.4 L = 1 mol

17.0 L =  0.759 moles of O2

Step 3: Calculate moles of S

For 1 mol of S we need 1 mol of O2 to produce 1 mol of SO2

For 0.759 moles of O2 we need 0.759 moles of S

Step 4: Calculate mass of S

Mass of S = moles S * molar mass S

Mass of S =  0.759 moles * 32.065 g/mol = 24.3 grams of S

5. By running a direct current through water, it can be decomposed it into its elements. How many liters of hydrogen gas can be produced from 1500 g of water?

Step 1: The balanced equation

2H2O → 2H2 + O2

Step 2: Calculate number of moles of water

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 1500 grams / 18.02 g/mol = 83.24 moles

Step 3: Calculate moles of H2

For 2 moles of H2O we'll have 2 moles of H2 and 1 mol of O2

For 83.24 moles of H2O we'll have 83.24 moles of H2

Step 4: Calculate volume of H2

1 mol = 22.4 L

83.24 moles = 22.4 * 83.24 = 1864.6 L of hydrogen gas

6. Hydrogen and chlorine can be combined to form hydrochloric acid. If you reacted 6  liters of hydrogen with chlorine how many grams of acid will be produced?

Step 1: The balanced equation

H2 + Cl2 → 2HCl

Step 2: Calculate moles of H2

22.4 L = 1 mol

6.0 L = 0.268 moles of H2 (Suppose H2 is the limiting reactant)

Step 3: Calculate moles of HCl

For 1 mol of H2 we need 1 mol of Cl2 to produce 2 moles of HCl

For 0.268 moles of H2 we'll have 2*0.268 = 0.536 moles of HCl

Step 4: Calculate mass of HCl

Mass HCl = moles HCl * molar mass HCl

Mass HCl = 0.536 * 36.46 g/mol

Mass HCl = 19.54 grams of HCl

7. . When glucose (C6H12O6) undergoes combustion with oxygen it produces water and carbon dioxide. How many liters of carbon dioxide will be produced if glucose reacts with 45 liters of oxygen?

Step 1: The balanced equation

C6H12O6 + 6O2 → 6CO2 + 6H2O

Step 2: Calculate moles of oxygen

22.4 L = 1 mol

45.0 L = 2.01 moles O2

Step 3: Calculate moles of CO2

For 1 mol of glucose we need 6 moles of O2 to prodcue 6 moles of CO2 and 6 moles of H2O

For 2.01 moles of O2 consumed we'll have 2.01 moles of CO2 produced

Step 4: Calculate liters of CO2 produced

1 mol = 22.4 L

2.01 moles = 45.0 L of CO2 produced

8. How many liters of nitrogen monoxide can be made using 18 liters of oxygen in the  following composition reaction?

Step 1: The balanced equation:

N2 + O2 → 2NO

Step 2: Calculate number of moles of O2

22.4 L = 1.0 mol

18.0 L = 0.804 moles O2

Step 3: Calculate moles of NO

For 1 mol of N2 we need 1 mol of O2 to produce 2 moles of NO

For 0.804 moles of O2 we need 2*0.804 = 1.608 moles of NO

Step 4: Calculate volume of NO

1 mol = 22.4 L

1.608 moles = 22.4 * 1.608 = 36.0 L of NO

9. How many liters of carbon dioxide are produced when 10 liters of propane (C3H8)  undergoes complete combustion? (assume STP)

Step 1: The balanced equation

C3H8 + 5O2 → 3CO2 + 4H2O

Step 2: Calculate moles of propane

22.4 L = 1.0 mol

10.0 L = 0.446 moles of propane

Step 3: Calculate moles of CO2

1 mol of propane consumed needs 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

For 0.446 moles of propane we'll have 3*0.446 = 1.338 moles of CO2

Step 4: Calculate volume of CO2

1 mol = 22.4 L

1.338 moles = 22.4 *1.338 = 30L of CO2

Chemistry

1. D (24.0 moles CO2)

2. A (.239 moles H2)

Explanation:

1. First Balance the equation

1 C3H8 + 5 O2 ---> 3 CO2 + 4 H2O

Then set up a stoiciometric equation so that the moles of O2 cancel out

40mol O2 x = 24.0 moles CO2

2. Set up a stoichiometric equation

10 grams Fe x x = 0.239 moles H2

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