Answer:
89.3 %
Explanation:
M(CH4) = 12+ 4*1 = 16 g/mol
M(O2) = 2*16 = 32 g/mol
M(CO2) = 12 + 2*16 = 44 g/mol
8.50 g * 1 mol/16 g = 0.5313 mol CH4
15.9 g * 1 mol/32 g = 0.4969 mol O2
9.77 g * 1 mol/44 g = 0.2220 mol CO2
1) CH4 + 2O2 -----> CO2 + 2H2O
from reaction 1 mol 2 mol
given 0.5313 mol (0.4969 mol)
1 mol CH4 --- 2 mol O2
0.5313 mol CH4 --- x mol O2
x= 2*0.5313 = 1.0626 mol O2
We can see that for given amount of CH4 we do not have enough O2, so O2 is a limiting reactant.
2) CH4 + 2O2 -----> CO2 + 2H2O
from reaction 2 mol 1 mol
given 0.4969 mol x mol
x = 0.4969*1/2 = 0.2485 mol CO2 theoretical yield
3)
Practical yield CO2 = 0.2220 mol
Theoretical yield CO2 = 0.2485 mol
% yield = (0.2220/0.2485)*100% = 89.3 %
