Question

Gaseous methane CH4 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Suppose 11. g of methane is mixed with 73.5 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 2 significant digits.

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)= 11.0 g

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(11.0 g)/(16.042 g/mol)

= 0.6857 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 73.5 g

number of mol of O2,

n = mass of O2/molar mass of O2

=(73.5 g)/(32 g/mol)

= 2.297 mol

Balanced chemical equation is:

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

1 mol of CH4 reacts with 2 mol of O2

for 0.6857 mol of CH4, 1.3714 mol of O2 is required

But we have 2.296875 mol of O2

so, CH4 is limiting reagent

we will use CH4 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 = (2/1)* moles of CH4

= (2/1)*0.6857

= 1.3714 mol

mass of H2O = number of mol * molar mass

= 1.371*18.02

= 24.71 g

Answer: 25 g