Question

If 50.0 g of methane react with 175.0 g of oxygen gas, how many liters of CO2 gas could be produced if the reaction was carried out at STP?

Answer 1

Molar mass of CH4,

MM = 1*MM(C) + 4*MM(H)

= 1*12.01 + 4*1.008

= 16.042 g/mol

mass(CH4)= 50.0 g

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(50.0 g)/(16.042 g/mol)

= 3.117 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 175.0 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(175.0 g)/(32 g/mol)

= 5.469 mol

Balanced chemical equation is:

CH4 + 2 O2 ---> CO2 + 2 H2O

1 mol of CH4 reacts with 2 mol of O2

for 3.116818 mol of CH4, 6.233637 mol of O2 is required

But we have 5.46875 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

According to balanced equation

mol of CO2 formed = (1/2)* moles of O2

= (1/2)*5.46875

= 2.73 mol

At STP, 1 mol of gas occupies 22.4 L

So,

volume =2.73 mol * 22.4 L/mol = 61.2 L

Answer: 61.2 L