What is the mass of 8 moles of Fe2O3 ?, №18010083, 13.01.2022 22:05

Chemistry

Step-by-step answer

P Answered by Specialist

What is the mass of 5 moles of Fe2O3 ?

798g

Chemistry

1. How many grams are there in 7.5 x 1023 molecules of H2SO4? 2. How many molecules are there in 122 grams of Cu(NO3)2? 3. How many moles in 28 grams of CO2? 4. What is the mass of 3 moles of Fe2O3? 5. How many moles of neon atoms are present in 16.3 L of neon gas at STP? 6. What is the volume 2 mol of chlorine Cl2 gas at STP 7. Find the mass in grams of 2.00 x 1023 molecules of F2. 8. Determine the volume in liters occupied by 14 g of nitrogen N2 gas at STP. 9. What is the relative density ofnitrogen oxide NO2 gas according to air? 10. What is

Step-by-step answer

P Answered by Specialist

I cannot give you all the answer but I can help you to solve those.

Explanation:

The first question:

How many grams are there in 7.5× $10^{23}$ molecules of $H_{2} SO_{4}$ ?

So we need to find the molecular mass first, use your periodic table,

And then we can find out 2+32+16×4=98 g/mol

Then, we need to find how many moles, by using Avogadro's constant:

Avogadro's constant: 1 mole = 6.02× $10^{23}$

∴ $\frac{7.5*10^{23} }{6.02*10^{23}}$ =1.25 mol(2d.p.)

Lastly, find the grams using the formula $M= \frac{m}{n}$

m=Mn

m=1.25*98

m=122.5g

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In conclusion, use those formula to help you:

$M= \frac{m}{n}$ (which M = molecular mass(atomic mass) m=mass of the substance and n = moles)

Avogadro's constant: $\frac {molecular mass} {6.02*10^{23}} = moles$

Chemistry

1. How many grams are there in 7.5 x 1023 molecules of H2SO4? 2. How many molecules are there in 122 grams of Cu(NO3)2? 3. How many moles in 28 grams of CO2? 4. What is the mass of 3 moles of Fe2O3? 5. How many moles of neon atoms are present in 16.3 L of neon gas at STP? 6. What is the volume 2 mol of chlorine Cl2 gas at STP 7. Find the mass in grams of 2.00 x 1023 molecules of F2. 8. Determine the volume in liters occupied by 14 g of nitrogen N2 gas at STP. 9. What is the relative density ofnitrogen oxide NO2 gas according to air? 10. What is

Step-by-step answer

P Answered by Master

I cannot give you all the answer but I can help you to solve those.

Explanation:

The first question:

How many grams are there in 7.5× $10^{23}$ molecules of $H_{2} SO_{4}$ ?

So we need to find the molecular mass first, use your periodic table,

And then we can find out 2+32+16×4=98 g/mol

Then, we need to find how many moles, by using Avogadro's constant:

Avogadro's constant: 1 mole = 6.02× $10^{23}$

∴ $\frac{7.5*10^{23} }{6.02*10^{23}}$ =1.25 mol(2d.p.)

Lastly, find the grams using the formula $M= \frac{m}{n}$

m=Mn

m=1.25*98

m=122.5g

-------------------------------------------------------------------------------------------------------------

In conclusion, use those formula to help you:

$M= \frac{m}{n}$ (which M = molecular mass(atomic mass) m=mass of the substance and n = moles)

Avogadro's constant: $\frac {molecular mass} {6.02*10^{23}} = moles$

Chemistry

85 points asap 1. how many molecules are in 7.5 grams of oxygen gas? 2. 2hgo(s) -- 2hg(l) + o2 (g) in a certain reaction 4.37 of hgo is decomposed, producing 2.5 of hg what is the percent yield for the reaction? 3. 4 fe(s) + 3o2(g) -- 2fe2o3(g) in a certain reaction, 27.3 g of iron is reacted with 79.9 g of oxygen how many grams of the excess reactant remain after the reaction is complete? 4. how many moles are in 10.25 grams of nh3? 5. what is the empirical formula mass (efm) of p2o5? 6. how many molecules are in 1.84 moles of water? 7. 4 fe(s)

Step-by-step answer

P Answered by PhD

1. Answer is: there are 1.41·10²³ molecules of oxygen.
1) calculate amount of substance for oxygen gas:
m(O₂) = 7.5 g; mass of oxygen.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 7.5 g ÷ 32 g/mol.
n(O₂) = 0.234 mol.
2) calculate number of molecules:
N(O₂) = n(O₂) · Na.
N(O₂) = 0.234 mol · 6.022·10²³ 1/mol.
N(O₂) = 1.41·10²³.
Na - Avogadro constant.

2. Answer is: the percent yield for the reaction is 61.77%.
Balanced chemical reaction: 2HgO(s) → 2Hg(l) + O₂(g).
m(HgO) = 4.37 g.
n(HgO) = n(HgO) ÷ M(HgO).
n(HgO) = 4.37 g ÷ 216.6 g/mol.
n(HgO) = 0.02 mol.
From chemical reaction: n(HgO) : n(Hg) = 2 : 2 (1 :1).
n(Hg) = n(HgO) = 0.02 mol; amount of substance.
m(Hg) = n(Hg) ·M(Hg).
m(Hg) = 0.02 mol · 200.6 g/mol.
m(Hg) = 4.047 g.
yield = 2.5 g ÷ 4.047 g · 100%.
yield = 61.77%.

3. Answer is: 68.16 grams of the excess reactant (oxygen) remain.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 79.9 g.
n(O₂) = 79.9 g ÷ 32 g/mol.
n(O₂) = 2.497 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
0.489 mol : n(O₂) = 4 : 3.
n(O₂) = 3 · 0.489 mol ÷ 4.
n(O₂) = 0.367 mol.
Δn(O₂) = 2.497 mol - 0.367 mol.
Δn(O₂) = 2.13 mol.
m(O₂) = 2.13 mol · 32 g/mol.
m(O₂) = 68.16 g.

4. Answer is: there are 0.603 moles of ammonia.
m(NH₃) = 10.25 g; mass of ammonia.
M(NH₃) = Ar(N) + 3Ar(H) · g/mol.
M(NH₃) = 14 + 3·1 · g/mol.
M(NH₃) = 17 g/mol; molar mass of ammonia.
n(NH₃) = m(NH₃) ÷ M(NH₃).
n(NH₃) = 10.25 g ÷ 17 g/mol.
n(NH₃) = 0.603 mol; amount of substance (ammonia).

5. Answer is: the empirical formula mass of P₂O₅ is 141.89.
Empirical formula gives the proportions of the elements present in a compound.
Atomic mass of phosphorus is 30.97 g/mol.
Atomic mass of oxygen is 15.99 g/mol.
In phosphorus (V) oxide there are two atoms of phosphorus and five atoms of oxygen:
EFM(P₂O₅) = 2·30.97 g/mol + 5·15.99 g/mol = 141.89 g/mol.

6. Answer is: there are 1.108·10²⁴ molecules of water.
n(H₂O) = 1.84 mol; amount of substance (water).
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 1.84 mol · 6.022·10²³ 1/mol.
N(H₂O) = 11.08·10²³.
N(H₂O) = 1.108·10²⁴.
Na - Avogadro constant (number of particles (ions, atoms or molecules), that are contained in one mole of substance).

7. Answer is: iron (Fe) is the limiting reactant.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 45.8 g.
n(O₂) = 45.8 g ÷ 32 g/mol.
n(O₂) = 1.431 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
For 1.431moles of oxygen we need:
1.431 mol : n(Fe) = 3 : 4.
n(Fe) = 1.908 mol, there is no enough iron.

8. Answer is: there are 0.435 moles of C₆H₁₄.
N(C₆H₁₄) = 2.62·10²³; number of molecules.
n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.
n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.
n(C₆H₁₄) = 0.435 mol; amount of substance of C₆H₁₄.
Na - Avogadro constant or Avogadro number.

9. Answer is: 3.675 moles of carbon(II) oxide are required to completely react.
Balanced chemical reaction: Fe₂O₃(s) + 3CO(g) ⟶ 2Fe(s) + 3CO₂(g).
n(Fe₂O₃) = 1.225 mol; amount of substance.
From chemical reaction: n(Fe₂O₃) : n(CO) = 1 : 3.
1.225 mol : n(CO) = 1 : 3.
n(CO) = 3 · 1.225 mol.
n(CO) = 3.675 mol.

10. Answer is: there are 2.158 moles of barium atoms.
N(Ba) = 2.62·10²³; number of atoms of barium.
n(Ba) = N(Ba) ÷ Na.
n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.
n(Ba) = 2.158 mol; amount of substance of barium.
Na - Avogadro constant or Avogadro number.

11. Answer is: 6.26·10²³ carbon atoms are present.
n(C₂H₆O) = 0.52 mol; amount of substance.
N(C₂H₆O) = n(C₂H₆O) · Na.
N(C₂H₆O) = 0.52 mol · 6.022·10²³ 1/mol.
N(C₂H₆O) = 3.13·10²³.
In one molecule of C₂H₆O there are two atoms of carbon:
N(C) = N(C₂H₆O) · 2.
N(C) = 3.13·10²³ · 2.
N(C) = 6.26·10²³.

12. Answer is: the empirical formula is C₂H₄O.
If we use 100 grams of compound:
1) ω(C) = 51% ÷ 100% = 0.51.
m(C) = ω(C) · m(compound).
m(C) = 0.51 · 100 g.
m(C) = 51 g.
n(C) = m(C) ÷ M(C).
n(C) = 51 g ÷ 12 g/mol.
n(C) = 4.25 mol.
2) ω(H) = 9.3 % ÷ 100% = 0.093.
m(H) = 0.093 · 100 g.
m(H) = 9.3 g.
n(H) = 9.3 g ÷ 1 g/mol.
n(H) = 9.3 mol
3) ω(O) = 39.2 % ÷ 100%.
ω(O) = 0.392.
m(O) = 0.392 · 100 g.
m(O) = 39.2 g.
n(O) = 39.2 g ÷ 16 g/mol.
n(O) = 2.45 mol.
4) n(C) : n(H) : n(O) = 4.25 mol : 9.3 mol : 2.45 mol /2.45 mol.
n(C) : n(H) : n(O) = 1.73 : 3.795 : 1.

Chemistry

85 points asap 1. how many molecules are in 7.5 grams of oxygen gas? 2. 2hgo(s) -- 2hg(l) + o2 (g) in a certain reaction 4.37 of hgo is decomposed, producing 2.5 of hg what is the percent yield for the reaction? 3. 4 fe(s) + 3o2(g) -- 2fe2o3(g) in a certain reaction, 27.3 g of iron is reacted with 79.9 g of oxygen how many grams of the excess reactant remain after the reaction is complete? 4. how many moles are in 10.25 grams of nh3? 5. what is the empirical formula mass (efm) of p2o5? 6. how many molecules are in 1.84 moles of water? 7. 4 fe(s)

Step-by-step answer

P Answered by PhD

1. Answer is: there are 1.41·10²³ molecules of oxygen.
1) calculate amount of substance for oxygen gas:
m(O₂) = 7.5 g; mass of oxygen.
n(O₂) = m(O₂) ÷ M(O₂).
n(O₂) = 7.5 g ÷ 32 g/mol.
n(O₂) = 0.234 mol.
2) calculate number of molecules:
N(O₂) = n(O₂) · Na.
N(O₂) = 0.234 mol · 6.022·10²³ 1/mol.
N(O₂) = 1.41·10²³.
Na - Avogadro constant.

2. Answer is: the percent yield for the reaction is 61.77%.
Balanced chemical reaction: 2HgO(s) → 2Hg(l) + O₂(g).
m(HgO) = 4.37 g.
n(HgO) = n(HgO) ÷ M(HgO).
n(HgO) = 4.37 g ÷ 216.6 g/mol.
n(HgO) = 0.02 mol.
From chemical reaction: n(HgO) : n(Hg) = 2 : 2 (1 :1).
n(Hg) = n(HgO) = 0.02 mol; amount of substance.
m(Hg) = n(Hg) ·M(Hg).
m(Hg) = 0.02 mol · 200.6 g/mol.
m(Hg) = 4.047 g.
yield = 2.5 g ÷ 4.047 g · 100%.
yield = 61.77%.

3. Answer is: 68.16 grams of the excess reactant (oxygen) remain.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 79.9 g.
n(O₂) = 79.9 g ÷ 32 g/mol.
n(O₂) = 2.497 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
0.489 mol : n(O₂) = 4 : 3.
n(O₂) = 3 · 0.489 mol ÷ 4.
n(O₂) = 0.367 mol.
Δn(O₂) = 2.497 mol - 0.367 mol.
Δn(O₂) = 2.13 mol.
m(O₂) = 2.13 mol · 32 g/mol.
m(O₂) = 68.16 g.

4. Answer is: there are 0.603 moles of ammonia.
m(NH₃) = 10.25 g; mass of ammonia.
M(NH₃) = Ar(N) + 3Ar(H) · g/mol.
M(NH₃) = 14 + 3·1 · g/mol.
M(NH₃) = 17 g/mol; molar mass of ammonia.
n(NH₃) = m(NH₃) ÷ M(NH₃).
n(NH₃) = 10.25 g ÷ 17 g/mol.
n(NH₃) = 0.603 mol; amount of substance (ammonia).

5. Answer is: the empirical formula mass of P₂O₅ is 141.89.
Empirical formula gives the proportions of the elements present in a compound.
Atomic mass of phosphorus is 30.97 g/mol.
Atomic mass of oxygen is 15.99 g/mol.
In phosphorus (V) oxide there are two atoms of phosphorus and five atoms of oxygen:
EFM(P₂O₅) = 2·30.97 g/mol + 5·15.99 g/mol = 141.89 g/mol.

6. Answer is: there are 1.108·10²⁴ molecules of water.
n(H₂O) = 1.84 mol; amount of substance (water).
N(H₂O) = n(H₂O) · Na.
N(H₂O) = 1.84 mol · 6.022·10²³ 1/mol.
N(H₂O) = 11.08·10²³.
N(H₂O) = 1.108·10²⁴.
Na - Avogadro constant (number of particles (ions, atoms or molecules), that are contained in one mole of substance).

7. Answer is: iron (Fe) is the limiting reactant.
Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).
m(Fe) = 27.3 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 27.3 g ÷ 55.85 g/mol.
n(Fe) = 0.489 mol.
m(O₂) = 45.8 g.
n(O₂) = 45.8 g ÷ 32 g/mol.
n(O₂) = 1.431 mol; amount of substance.
From chemical reaction: n(Fe) . n(O₂) = 4 : 3.
For 1.431moles of oxygen we need:
1.431 mol : n(Fe) = 3 : 4.
n(Fe) = 1.908 mol, there is no enough iron.

8. Answer is: there are 0.435 moles of C₆H₁₄.
N(C₆H₁₄) = 2.62·10²³; number of molecules.
n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.
n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.
n(C₆H₁₄) = 0.435 mol; amount of substance of C₆H₁₄.
Na - Avogadro constant or Avogadro number.

9. Answer is: 3.675 moles of carbon(II) oxide are required to completely react.
Balanced chemical reaction: Fe₂O₃(s) + 3CO(g) ⟶ 2Fe(s) + 3CO₂(g).
n(Fe₂O₃) = 1.225 mol; amount of substance.
From chemical reaction: n(Fe₂O₃) : n(CO) = 1 : 3.
1.225 mol : n(CO) = 1 : 3.
n(CO) = 3 · 1.225 mol.
n(CO) = 3.675 mol.

10. Answer is: there are 2.158 moles of barium atoms.
N(Ba) = 2.62·10²³; number of atoms of barium.
n(Ba) = N(Ba) ÷ Na.
n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.
n(Ba) = 2.158 mol; amount of substance of barium.
Na - Avogadro constant or Avogadro number.

11. Answer is: 6.26·10²³ carbon atoms are present.
n(C₂H₆O) = 0.52 mol; amount of substance.
N(C₂H₆O) = n(C₂H₆O) · Na.
N(C₂H₆O) = 0.52 mol · 6.022·10²³ 1/mol.
N(C₂H₆O) = 3.13·10²³.
In one molecule of C₂H₆O there are two atoms of carbon:
N(C) = N(C₂H₆O) · 2.
N(C) = 3.13·10²³ · 2.
N(C) = 6.26·10²³.

12. Answer is: the empirical formula is C₂H₄O.
If we use 100 grams of compound:
1) ω(C) = 51% ÷ 100% = 0.51.
m(C) = ω(C) · m(compound).
m(C) = 0.51 · 100 g.
m(C) = 51 g.
n(C) = m(C) ÷ M(C).
n(C) = 51 g ÷ 12 g/mol.
n(C) = 4.25 mol.
2) ω(H) = 9.3 % ÷ 100% = 0.093.
m(H) = 0.093 · 100 g.
m(H) = 9.3 g.
n(H) = 9.3 g ÷ 1 g/mol.
n(H) = 9.3 mol
3) ω(O) = 39.2 % ÷ 100%.
ω(O) = 0.392.
m(O) = 0.392 · 100 g.
m(O) = 39.2 g.
n(O) = 39.2 g ÷ 16 g/mol.
n(O) = 2.45 mol.
4) n(C) : n(H) : n(O) = 4.25 mol : 9.3 mol : 2.45 mol /2.45 mol.
n(C) : n(H) : n(O) = 1.73 : 3.795 : 1.

Chemistry

Guys please answer this Fe2O3+3C — 2Fe +3CO If we took 8 moles of carbon how many grams(mass) it would be??

Step-by-step answer

P Answered by PhD

1

Mass = 96 g

Explanation:

Given data:

Number of moles of C = 8 mol

Mass of C in gram = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of C = 12 g/mol

8 mol = mass / 12 g/mol

Mass = 8 mol × 12 g/mol

Mass = 96 g

Chemistry

Guys please answer this Fe2O3+3C — 2Fe +3CO If we took 8 moles of carbon how many grams(mass) it would be??

Step-by-step answer

P Answered by PhD

1

Mass = 96 g

Explanation:

Given data:

Number of moles of C = 8 mol

Mass of C in gram = ?

Solution:

Formula:

Number of moles = mass/molar mass

Molar mass of C = 12 g/mol

8 mol = mass / 12 g/mol

Mass = 8 mol × 12 g/mol

Mass = 96 g

Chemistry

Question 1(multiple choice worth 4 points) (05.04 mc) what mass of co is needed to react completely with 55.0 g of fe2o3 in the reaction: fe2o3(s) + co(g) → fe(s) + co2(g)? 4.82 g co 9.64 g co 14.5 g co 28.9 g co question 2(multiple choice worth 4 points) (05.04 lc) which of the following is a valid mole ratio from the balanced equation 2c3h6 + 9o2 → 6co2 + 6h2o? one mole of c three h six over two moles of c o two six moles of h two o over nine moles of o two two moles of c three h six over six moles of o two three moles of h two o over 2 moles

Step-by-step answer

P Answered by PhD

76

Answer

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of $H_{2}O$ over Nine moles of $O_{2}$

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Answer 4 : Mass of $O_{2}$ = (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 $Fe_{2}O_{3}$ = 55 g

Molar mass of $Fe_{2}O_{3}$ = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of $Fe_{2}O_{3}$ = $\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}$ = $\frac{55 g}{159.69 g/mole}$ = 0.344 moles

Balanced chemical reaction is,

$Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)$

From the given reaction, we conclude that

1 mole of $Fe_{2}O_{3}$ gives → 3 moles of CO

0.344 moles of $Fe_{2}O_{3}$ 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,

$2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O$

From the given reaction, we conclude that the Six moles of $H_{2}O$ over Nine moles of $O_{2}$ is the correct option.

Solution 3 : The balanced chemical reaction is,

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

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 $Fe_{2}O_{3}$ .

Solution 4 : Given,

Given mass of $Zn(ClO_{3})_{2}$ = 150 g

Molar mass of $Zn(ClO_{3})_{2}$ = 232.29 g/mole

Molar mass of $O_{2}$ = 31.998 g/mole

Moles of $Zn(ClO_{3})_{2}$ = $\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}$ = $(\frac{150\times 1}{232.29})moles$

The balanced chemical equation is,

$Zn(ClO_{3})_{2}}\rightarrow ZnCl_{2}+3O_{2}$

From the given balanced equation, we conclude that

1 mole of $Zn(ClO_{3})_{2}$ gives → 3 moles of $O_{2}$

$(\frac{150\times 1}{232.29})moles$ of $Zn(ClO_{3})_{2}$ gives → $[(\frac{150\times 1}{232.29})\times 3] moles$ of $O_{2}$

Mass of $O_{2}$ = Number of moles of $O_{2}$ × Molar mass of $O_{2}$ = $[(\frac{150\times 1}{232.29})\times 3] \times 31.998 grams$

Therefore, the mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of $Na_{2}SO_{4}$ = 4.2 moles

Balanced chemical equation is,

$H_{2}SO_{4}+2NaCN\rightarrow 2HCN+Na_{2}SO_{4}$

From the given chemical reaction, we conclude that

1 mole of $Na_{2}SO_{4}$ obtained from 2 moles of NaCN

4.2 moles of $Na_{2}SO_{4}$ obtained → 2 × 4.2 moles of NaCN

Therefore,

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

Chemistry

Question 1(multiple choice worth 4 points) (05.04 mc) what mass of co is needed to react completely with 55.0 g of fe2o3 in the reaction: fe2o3(s) + co(g) → fe(s) + co2(g)? 4.82 g co 9.64 g co 14.5 g co 28.9 g co question 2(multiple choice worth 4 points) (05.04 lc) which of the following is a valid mole ratio from the balanced equation 2c3h6 + 9o2 → 6co2 + 6h2o? one mole of c three h six over two moles of c o two six moles of h two o over nine moles of o two two moles of c three h six over six moles of o two three moles of h two o over 2 moles

Step-by-step answer

P Answered by PhD

76

Answer

Answer 1 : 28.9 g of CO is needed.

Answer 2 : Six moles of $H_{2}O$ over Nine moles of $O_{2}$

Answer 3 : Four over two fraction can be used for the mole ratio to determine the mass of Fe from a known mass of $Fe_{2}O_{3}$ .

Answer 4 : Mass of $O_{2}$ = (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 $Fe_{2}O_{3}$ = 55 g

Molar mass of $Fe_{2}O_{3}$ = 159.69 g/mole

Molar mass of CO = 28.01 g/mole

Moles of $Fe_{2}O_{3}$ = $\frac{\text{ Given mass of } Fe_{2}O_{3}}{\text{ Molar mass of } Fe_{2}O_{3}}$ = $\frac{55 g}{159.69 g/mole}$ = 0.344 moles

Balanced chemical reaction is,

$Fe_{2}O_{3}(s)+3CO(g)\rightarrow 2Fe(s)+3CO_{2}(g)$

From the given reaction, we conclude that

1 mole of $Fe_{2}O_{3}$ gives → 3 moles of CO

0.344 moles of $Fe_{2}O_{3}$ 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,

$2C_{3}H_{6}+9O_{2}\rightarrow 6CO_{2}+6H_{2}O$

From the given reaction, we conclude that the Six moles of $H_{2}O$ over Nine moles of $O_{2}$ is the correct option.

Solution 3 : The balanced chemical reaction is,

$4Fe+3O_{2}\rightarrow 2Fe_{2}O_{3}$

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 $Fe_{2}O_{3}$ .

Solution 4 : Given,

Given mass of $Zn(ClO_{3})_{2}$ = 150 g

Molar mass of $Zn(ClO_{3})_{2}$ = 232.29 g/mole

Molar mass of $O_{2}$ = 31.998 g/mole

Moles of $Zn(ClO_{3})_{2}$ = $\frac{\text{ Given mass of }Zn(ClO_{3})_{2} }{\text{ Molar mass of } Zn(ClO_{3})_{2}}$ = $(\frac{150\times 1}{232.29})moles$

The balanced chemical equation is,

$Zn(ClO_{3})_{2}}\rightarrow ZnCl_{2}+3O_{2}$

From the given balanced equation, we conclude that

1 mole of $Zn(ClO_{3})_{2}$ gives → 3 moles of $O_{2}$

$(\frac{150\times 1}{232.29})moles$ of $Zn(ClO_{3})_{2}$ gives → $[(\frac{150\times 1}{232.29})\times 3] moles$ of $O_{2}$

Mass of $O_{2}$ = Number of moles of $O_{2}$ × Molar mass of $O_{2}$ = $[(\frac{150\times 1}{232.29})\times 3] \times 31.998 grams$

Therefore, the mass of $O_{2}$ = (150 × 3 × 31.998) ÷ (232.29 × 1) grams

Solution 5 : Given,

Number of moles of $Na_{2}SO_{4}$ = 4.2 moles

Balanced chemical equation is,

$H_{2}SO_{4}+2NaCN\rightarrow 2HCN+Na_{2}SO_{4}$

From the given chemical reaction, we conclude that

1 mole of $Na_{2}SO_{4}$ obtained from 2 moles of NaCN

4.2 moles of $Na_{2}SO_{4}$ obtained → 2 × 4.2 moles of NaCN

Therefore,

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

Chemistry

Me given the reaction, pcl3 + cl2 imported asset pcl5, if 3.00 moles of cl2 are used, then how many moles of pcl5 are made? 6.00 moles 5.00 moles 3.00 moles 1.00 mole given the reaction, fe2o3 + 3 co imported asset 2 fe + 3 co2, if 12.0 moles of co2 are made, then how many moles of co were used? 12.0 moles 6.0 moles 4.0 moles 3.0 moles given the reaction, 2 so2 + o2 imported asset 2 so3, if 4 moles of so2 are used, then how many moles of so3 are made? 4 moles 2 moles 1 mole 1/2 mole using the equation, 4fe + 3o2 imported asset 2fe2o3, if 6 moles

Step-by-step answer

P Answered by Master

10

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.

What is the mass of 8 moles of Fe2O3 ?

Step-by-step answer

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