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)

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

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

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

the correct answer is:

a. formation of precipitation