Question

CH_{4 (g) +} O₂(g) +H₂O-------> CO₂ (g) + H₂O

What would the percent yield of the reaction be if 15.2 g of O₂ were reacted and only 6.7 g of CO₂ were made?

Answer 1

Given the mass of O2 reacted = 15.2 g

Molecular mass of O2 = 32.0 g/mol

Hence moles of O2 reacted = mass / molecular mass = 15.2 g / 32.0 g/mol = 0.475 mol O2

Given the mass of CO2 produced = 6.7 g

Molecular mass of CO2 = 44.0 g/mol

Hence moles of CO2 produced = mass / molecular mass = 6.7 g / 44.0 g/mol = 0.1523 mol CO2

Actually here the proper balanced reaction for the combustio of methane is

CH4 (g) + 2 O2(g) -------> CO2 (g) + 2 H2O

1 mol ----- 2 mol ------------1 mol ------ 2 mol

From the above balanced reaction it is clear that 2 moles of O2 produces 1 mole of CO2.

Hence 0.475 mol O2 that will produce the stoichiometric moles of CO2 = 0.475 mol O2 x (1 mol CO2 / 2 mol O2)

= 0.2375 mol CO2

Hence theoritical mass of CO2 produced = 0.2375 mol CO2 x (44 g CO2 / 1 mol CO2) = 10.45 g CO2

However actual mass of CO2 produced = 6.7 g

Hence percent yield = (actual yield / theoritical yield) x 100 = (6.7 g / 10.45 g ) x 100 = 64.1 % (answer)