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

i'm not sure

explanation:

i'll try whatever you got

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current