Question

Gaseous methane CH4 reacts with gaseous oxygen gas O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O . Is 1.88g of water is produced from the reaction of 0.96g of methane and 7.2g of oxygen gas, calculate the percent yield of water. Round your answer to 2 significant figures.

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)= 0.96 g

use:

number of mol of CH4,

n = mass of CH4/molar mass of CH4

=(0.96 g)/(16.04 g/mol)

= 5.984*10^-2 mol

Molar mass of O2 = 32 g/mol

mass(O2)= 7.2 g

use:

number of mol of O2,

n = mass of O2/molar mass of O2

=(7.2 g)/(32 g/mol)

= 0.225 mol

Balanced chemical equation is:

2 CH4 + 3 O2 ---> 4 H2O + 2 CO2

2 mol of CH4 reacts with 3 mol of O2

for 5.984*10^-2 mol of CH4, 8.976*10^-2 mol of O2 is required

But we have 0.225 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 = (4/2)* moles of CH4

= (4/2)*5.984*10^-2

= 0.1197 mol

use:

mass of H2O = number of mol * molar mass

= 0.1197*18.02

= 2.156 g

% yield = actual mass*100/theoretical mass

= 1.88*100/2.156

= 87.2 %

Answer: 87 %