Question

How many liters of CH4 are needed to produce 5.46 kg of H2 using the steam reformation process?

Data:


P = 2280 mmHg

T = 700 degrees Celsius


Steam Reformation:


CH4 (g) + H2O (g) = CO (g) + 3H2 (g)

Answer 1

2.40 × 10³ L

Step-by-step explanation:

You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.

M_r: 2.016

CH₄ + H₂O ⟶ CO + 3H₂

m/g: 5460

Step 1. Convert grams of H₂ to moles of H₂

1 mol H₂ = 2.016 g H₂

Moles of H₂ = 5460 × 1/2.016

Moles of H₂ = 2708 mol H₂

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Step 2. Convert moles of H₂ to moles of CH₄.

The molar ratio is 1 mol CH₄ to 3 mol H₂

Moles of CH₄ = 2708 × 1/3

Moles of CH₄ = 902.8 mol CH₄

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Step 3. Use the Ideal Gas Law to calculate the volume.

pV = nRT Divide each side by V

V = (nRT)/p

n = 902.8 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = (700 + 273.15) K = 973.15 K

p = 2280 mmHg × 1 atm/760 mmHg = 3.000 atm

V = (902.8 × 0.082 06 × 973.15)/3.000

V = 72 090/3.000

V = 2.40 × 10³ L