Question

When 14.363 g of Ga reacted with an excess of oxygen, a total of 9.656 g of Ga₂O₃ formed. What was the percent yield? Please break down the answer step by step.

Answer 1

Molar mass of Ga = 69.72 g/mol


mass(Ga)= 14.363 g

number of mol of Ga,
n = mass of Ga/molar mass of Ga
=(14.363 g)/(69.72 g/mol)
= 0.206 mol
Balanced chemical equation is:
4 Ga + 3 O2 ---> 2 Ga2O3 +


Molar mass of Ga2O3,
MM = 2*MM(Ga) + 3*MM(O)
= 2*69.72 + 3*16.0
= 187.44 g/mol

According to balanced equation
mol of Ga2O3 formed = (2/4)* moles of Ga
= (2/4)*0.20601
= 0.103005 mol


mass of Ga2O3 = number of mol * molar mass
= 0.103*1.874*10^2
= 19.31 g

% yield = actual mass*100/theoretical mass
= 9.656*100/19.307234
= 50.01 %
Answer: 50.01 %