Question

Consider the unbalanced reaction NiS2(s) + O2(g) --> NiO(s) + SO2(g). When 11.2g of NiS2 react with 5.43g of O2, 4.86g of NiO are obtained. The theoretical yield of Nio is? The limiting reactant is? The percentage yield is what percent?

Answer 1

Molar mass of NiS2,
MM = 1*MM(Ni) + 2*MM(S)
= 1*58.69 + 2*32.07
= 122.83 g/mol


mass(NiS2)= 11.2 g

number of mol of NiS2,
n = mass of NiS2/molar mass of NiS2
=(11.2 g)/(122.83 g/mol)
= 9.118*10^-2 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 5.43 g

number of mol of O2,
n = mass of O2/molar mass of O2
=(5.43 g)/(32 g/mol)
= 0.1697 mol
Balanced chemical equation is:
2 NiS2 + 5 O2 ---> 2 NiO + 4 SO2


2 mol of NiS2 reacts with 5 mol of O2
for 0.091183 mol of NiS2, 0.227957 mol of O2 is required
But we have 0.169687 mol of O2

so, O2 is limiting reagent
we will use O2 in further calculation


Molar mass of NiO,
MM = 1*MM(Ni) + 1*MM(O)
= 1*58.69 + 1*16.0
= 74.69 g/mol

According to balanced equation
mol of NiO formed = (2/5)* moles of O2
= (2/5)*0.169687
= 0.067875 mol


mass of NiO = number of mol * molar mass
= 6.787*10^-2*74.69
= 5.07 g

% yield = actual mass*100/theoretical mass
= 4.86*100/5.07
= 95.9 %
Answer:
Theoretical yield = 5.07 g
limiting reagent : NiS2
percentage yield = 95.9 %