Question

The percent yield of chromium(III) carbonate is 34.7 %, when chromium(III) nitrate reacts with sodium carbonate shown in the unbalanced reaction below. What is the amount of sodium carbonate required to react with excess chromium(III) nitrate to produce 45.0 g of chromium(III) carbonate.

Na₂CO₃(aq) + Cr(NO₃)₃(aq) → NaNO₃(aq) + Cr₂(CO₃­)₃(s)

Answer 1

% yield = actual yield * 100 / theoretical yield

34.7 = 45.0*100/theoretical yield

theoretical yield = 129.7 g

mass of Cr2(CO3)3 = 129.7 g

molar mass of Cr2(CO3)3 = 284.03 g/mol

mol of Cr2(CO3)3 = (mass)/(molar mass)

= 129.7/284.03

= 0.456642 mol

Balanced chemical equation is:

3 Na2CO3(aq) + 2 Cr(NO3)3(aq) → 6 NaNO3(aq) + Cr2(CO3­)3

According to balanced equation

mol of Na2CO3 required = (3/1)* moles of Cr2(CO3)3

= (3/1)*0.456642

= 1.369926 mol

mass of Na2CO3 = number of mol * molar mass

= 1.369926*105.99

= 145 g

Answer: 145 g