Question

Gaseous methane CH4 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 0.32 g of methane is mixed with 0.909 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction.

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)= 0.32 g

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(0.32 g)/(16.042 g/mol)

= 1.995*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 0.909 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(0.909 g)/(32 g/mol)

= 2.841*10^-2 mol

Balanced chemical equation is:

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

1 mol of CH4 reacts with 2 mol of O2

for 0.019948 mol of CH4, 0.039895 mol of O2 is required

But we have 0.028406 mol of O2

so, O2 is limiting reagent

we will use O2 in further calculation

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

According to balanced equation

mol of H2O formed = (1/2)* moles of O2

= (1/2)*0.028406

= 0.014203 mol

mass of H2O = number of mol * molar mass

= 1.42*10^-2*18.02

= 0.256 g

Answer: 0.256 g