Question

What volume of H2 gas (in L), measured at 742 mmHg and 90 ∘C, is required to synthesize 27.8 g CH3OH?

How many liters of CO gas, measured under the same conditions, is required?

Answer 1

27.8g of CH3OH is = 27.8/32 moles=0.86875 mol

hence we know 2 moles of H2 are required for one mole of CH3OH hence moles of H2 required for 0.86875 mol =0.86875*2=1.7375 mol

Using PV=nRT where P=742 mm =742/760 atm=0.976 atm,R=0.0821 T=90+273=363K,n=1.7375

We can calculate V=(1.7375*0.0821*363)/0.976=53.05 L

For part 2

1 mole of CO required for 1 mole of CH3OH

hence moles of CO required for 0.86875 mol = 0.86875 mol

Using PV=nRT where P=742 mm =742/760 atm=0.976 atm,R=0.0821 T=90+273=363K,n=0.86875

Hence V_CO=(0.86875*0.0821*363)/0.976=26.527 L

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