How much hydrogen gas is necessary to exert a pressure of 1.4 ATM at 430k if occupying a volume of 15.1 l

0.60 mol H₂

General Formula and Concepts:

Chemistry - Gas Laws

Ideal Gas Law: PV = nRT

Explanation:

Step 1: Define Variables

P = 1.4 atm

T = 430 K

R = 0.0821 (L · atm)/(mol · K)

V = 15.1 L

n = unknown

Step 2: Solve for n

Step 3: Check

We are given 2 sig figs as our lowest. Follow sig fig rules and round.

0.598816 mol H₂ ≈ 0.60 mol H₂

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as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2