350 g of barium nitrate reacts with hydrochloric acid to form 60 g of varium chloride.

350 g of barium nitrate reacts with hydrochloric acid to form 60 g of varium chloride. Determine the percentage of barium nitrate impurities.

Given:
m (Ba (NO3) 2) = 350 g
m (BaCl2) = 60 g

To find:
ω approx. -?

Decision:
1) Ba (NO3) 2 + 2HCl => BaCl2 + 2HNO3;
2) M (BaCl2) = Mr (BaCl2) = Ar (Ba) * N (Ba) + Ar (Cl) * N (Cl) = 137 * 1 + 35.5 * 2 = 208 g / mol;
M (Ba (NO3) 2) = Mr (Ba (NO3) 2) = Ar (Ba) * N (Ba) + Ar (N) * N (N) + Ar (O) * N (O) = 137 * 1 + 14 * 2 + 16 * 3 * 2 = 261 g / mol;
3) n (BaCl2) = m (BaCl2) / M (BaCl2) = 60/208 = 0.29 mol;
4) n (Ba (NO3) 2) = n (BaCl2) = 0.29 mol;
5) m clean. (Ba (NO3) 2) = n (Ba (NO3) 2) * M (Ba (NO3) 2) = 0.29 * 261 = 75.69 g;
6) m approx. = m (Ba (NO3) 2) – m pure. (Ba (NO3) 2) = 350 – 75.69 = 274.31 g;
7) ω approx. = m approx. * 100% / m (Ba (NO3) 2) = 274.31 * 100% / 350 = 78.37%.

Answer: The mass fraction of impurities is 78.37%.