How many grams of hydrogen are in 46 g of CH40?, №18010084, 13.01.2022 22:05

Chemistry

how many grams of hydrogen are in 46 g of ch4oh? answer is 6.97

Step-by-step answer

P Answered by Master

4

No! The answer is 5.75 g
Below is the solution.
how many grams of hydrogen are in 46 g of ch4oh? answer is 6.97

Chemistry

How many grams of hydrogen are in 46 g of CH40?

Step-by-step answer

P Answered by Specialist

mass H =

$\tt \dfrac{molar~mass~H}{molar~mass~CH_4O}\times mass~CH_4O\\\\\dfrac{4}{32}\times 46~g=5.75~g$

Chemistry

how many grams of hydrogen are in 46 g of ch4oh? answer is 6.97

Step-by-step answer

P Answered by Master

4

No! The answer is 5.75 g
Below is the solution.
how many grams of hydrogen are in 46 g of ch4oh? answer is 6.97

Chemistry

Nitrogen and hydrogen react together to form ammonia according to the equation: (g) + (g) → (g) (unbalanced)balance the equation, and determine how many grams of hydrogen would be required to form 50.0 g of ammonia, assuming there is sufficient nitrogen available.4.46 g5.94 g4.81 g8.91 gnot enough information

Step-by-step answer

P Answered by Specialist

$m_{H}=8.89 g$

Explanation:

To balance the equation:

1) Asume 1 mol of N2 :

$N_2 (g) + __ H_2 (g) \longrightarrow __ NH_3 (g)$

2) Balance the nitrogen:

$N_2 (g) + __ H_2 (g) \longrightarrow 2 NH_3 (g)$

3) Balance the hydrogen:

$N_2 (g) + 3 H_2 (g) \longrightarrow 2 NH_3 (g)$

To calculate the hydrogen:

$M_{am}= molecular weight of ammonia$

$M_{am}= 14+3*1=17 g/mol$

$n_{am}= mol of ammonia$

$n_{am}= \frac{50g}{17 g/mol}=2.94 mol$

Due to the equation you need 3 mol of hydrogen per 2 mol of ammonia:

$n_{H}= 2.94 mol \frac{3}{2}=4.41 mol$

$m_{H}=2.016 \frac{g}{mol}*4.41 mol=8.89 g$

Chemistry

Nitrogen and hydrogen react together to form ammonia according to the equation: (g) + (g) → (g) (unbalanced)balance the equation, and determine how many grams of hydrogen would be required to form 50.0 g of ammonia, assuming there is sufficient nitrogen available.4.46 g5.94 g4.81 g8.91 gnot enough information

Step-by-step answer

P Answered by Master

$m_{H}=8.89 g$

Explanation:

To balance the equation:

1) Asume 1 mol of N2 :

$N_2 (g) + __ H_2 (g) \longrightarrow __ NH_3 (g)$

2) Balance the nitrogen:

$N_2 (g) + __ H_2 (g) \longrightarrow 2 NH_3 (g)$

3) Balance the hydrogen:

$N_2 (g) + 3 H_2 (g) \longrightarrow 2 NH_3 (g)$

To calculate the hydrogen:

$M_{am}= molecular weight of ammonia$

$M_{am}= 14+3*1=17 g/mol$

$n_{am}= mol of ammonia$

$n_{am}= \frac{50g}{17 g/mol}=2.94 mol$

Due to the equation you need 3 mol of hydrogen per 2 mol of ammonia:

$n_{H}= 2.94 mol \frac{3}{2}=4.41 mol$

$m_{H}=2.016 \frac{g}{mol}*4.41 mol=8.89 g$

Chemistry

How many grams are in a mole of hydrogen chloride (hcl)? 1 h 1.01 hydrogen 17 cl 35.45 chlorine a. 1.5 grams/mole b. 35.45 grams/mole c. 36.46 grams/mole d. 72.92 grams/mole

Step-by-step answer

P Answered by Master

9

The molecular mass is the sum of the atomic masses of the atoms.

Atomic mass :

H = 1.01 amu

Cl = 35.45 amu

HCl = 1.01 + 35.45

HCl = 36.46 grams/mole

Therefore:

1 mole of HCl is equivalent to 36.46 grams/ mole

answer C

Chemistry

How many grams are in a mole of hydrogen chloride (hcl)? 1 h 1.01 hydrogen 17 cl 35.45 chlorine a. 1.5 grams/mole b. 35.45 grams/mole c. 36.46 grams/mole d. 72.92 grams/mole

Step-by-step answer

P Answered by Specialist

9

The molecular mass is the sum of the atomic masses of the atoms.

Atomic mass :

H = 1.01 amu

Cl = 35.45 amu

HCl = 1.01 + 35.45

HCl = 36.46 grams/mole

Therefore:

1 mole of HCl is equivalent to 36.46 grams/ mole

answer C

Chemistry

1. an unknown amount of mercury (ii) oxide was decomposed in the lab. mercury metal was formed and 4.50 l of oxygen gas was released at a pressure of 0.970 atm and 390.0 k. what was the initial weight of mercury oxide in the sample? a.) 27.3 grams b.) 38.3 grams c.) 46.2 grams d.) 60.6 grams 2. read the chemical equation. mg + hcl → mgcl2 + h2 how many liters of hydrogen gas is produced at 298 k and 0.960 atm if 3.00 moles of hydrochloric acid react with an excess of magnesium metal? a.) 38.2 liters b.) 42.6 liters c.) 66.7 liters d.) 76.4 liters

Step-by-step answer

P Answered by PhD

9

1. 59.1 g HgO

2. A.) 38.2 L

3. B.) It is directly proportional to the number of moles of the gas.

Step-by-step explanation:

1. Mass of HgO

We know we will need a chemical equation with masses and molar masses, so let's gather all the information in one place.

M_r: 216.59

2HgO ⟶ 2Hg + O₂

(a) Calculate the moles of O₂

pV = nRT Divide each side by RT

n = (pV)/(RT)

Data:

p = 0.970 atm

V = 4.50 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 390.0 K

Calculation:

n = (0.970 × 4.500)/(0.082 06 × 390.0)

n = 0.1364 mol O₂

(b) Calculate the moles of HgO

The molar ratio is 1 mol O₂/2 mol HgO.

Moles of HgO = 0.1364 mol O₂ × (2 mol Hg/1 mol O₂)

Moles of HgO = 0.2728 mol HgO

(c) Calculate the mass of HgO

Mass of HgO = 0.2728 mol HgO × (216.59 g HgO/1 mol HgO)

Mass of HgO = 59.1 g HgO

This isn't any of the options in your list. Have I made an error?

2. Volume of Hydrogen

Mg + 2HCl ⟶ MgCl₂ + H₂

n/mol: 3.00

(a) Calculate the moles of H₂

The molar ratio is (1 mol H₂/2 mol HCl).

Moles of H₂ = 3.00 mol HCl × (1 mol H₂/2 mol HCl)

Moles of H₂ = 1.50 mol HCl

(b) Calculate the volume of H₂

pV = nRT Divide both sides by p

V = (nRT)/p

Data:

n = 1.50 mol

T = 298 K

p = 0.960 atm

Calculation:

V = (1.5 × 0.082 06 × 298)/0.960

V = 38.2 L

3. Volume of gas

pV = nRT

V = nRT/p

If T and p are constant, (RT)/p = constant = k

V = kn

V ∝ n

The volume is directly proportional to the number of moles of the gas.

Chemistry

1. an unknown amount of mercury (ii) oxide was decomposed in the lab. mercury metal was formed and 4.50 l of oxygen gas was released at a pressure of 0.970 atm and 390.0 k. what was the initial weight of mercury oxide in the sample? a.) 27.3 grams b.) 38.3 grams c.) 46.2 grams d.) 60.6 grams 2. read the chemical equation. mg + hcl → mgcl2 + h2 how many liters of hydrogen gas is produced at 298 k and 0.960 atm if 3.00 moles of hydrochloric acid react with an excess of magnesium metal? a.) 38.2 liters b.) 42.6 liters c.) 66.7 liters d.) 76.4 liters

Step-by-step answer

P Answered by PhD

9

1. 59.1 g HgO

2. A.) 38.2 L

3. B.) It is directly proportional to the number of moles of the gas.

Step-by-step explanation:

1. Mass of HgO

We know we will need a chemical equation with masses and molar masses, so let's gather all the information in one place.

M_r: 216.59

2HgO ⟶ 2Hg + O₂

(a) Calculate the moles of O₂

pV = nRT Divide each side by RT

n = (pV)/(RT)

Data:

p = 0.970 atm

V = 4.50 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 390.0 K

Calculation:

n = (0.970 × 4.500)/(0.082 06 × 390.0)

n = 0.1364 mol O₂

(b) Calculate the moles of HgO

The molar ratio is 1 mol O₂/2 mol HgO.

Moles of HgO = 0.1364 mol O₂ × (2 mol Hg/1 mol O₂)

Moles of HgO = 0.2728 mol HgO

(c) Calculate the mass of HgO

Mass of HgO = 0.2728 mol HgO × (216.59 g HgO/1 mol HgO)

Mass of HgO = 59.1 g HgO

This isn't any of the options in your list. Have I made an error?

2. Volume of Hydrogen

Mg + 2HCl ⟶ MgCl₂ + H₂

n/mol: 3.00

(a) Calculate the moles of H₂

The molar ratio is (1 mol H₂/2 mol HCl).

Moles of H₂ = 3.00 mol HCl × (1 mol H₂/2 mol HCl)

Moles of H₂ = 1.50 mol HCl

(b) Calculate the volume of H₂

pV = nRT Divide both sides by p

V = (nRT)/p

Data:

n = 1.50 mol

T = 298 K

p = 0.960 atm

Calculation:

V = (1.5 × 0.082 06 × 298)/0.960

V = 38.2 L

3. Volume of gas

pV = nRT

V = nRT/p

If T and p are constant, (RT)/p = constant = k

V = kn

V ∝ n

The volume is directly proportional to the number of moles of the gas.

Chemistry

How many grams of xef6 are required to react with 0.579 l of hydrogen gas at 6.46 atm and 45°c in the reaction shown below?

Step-by-step answer

P Answered by PhD

1

The chemical reaction equation for this is

XeF6 + 3H2 ---> Xe + 6HF

Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.

we get n = 0.1433 moles H2

to get the mass of XeF6,

we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.

0.1433 moles H2 x $\frac{1 mole XeF6}{3 moles H2}$ x $\frac{245.28 g XeF6}{1 mole XeF6}$ = 11.71 g XeF6

Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.

How many grams of hydrogen are in 46 g of CH40?

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