14 g of calcium oxide was treated with a solution containing 36 g of nitric acid. What is the mass of the salt obtained?

Given:
m (CaO) = 14 g
m (HNO3) = 36 g

To find:
m (salt) -?

Decision:
1) СaO + 2HNO3 => Ca (NO3) 2 + H2O;
2) M (CaO) = Mr (CaO) = Ar (Ca) * N (Ca) + Ar (O) * N (O) = 40 * 1 + 16 * 1 = 56 g / mol;
M (HNO3) = Mr (HNO3) = Ar (H) * N (H) + Ar (N) * N (N) + Ar (O) * N (O) = 1 * 1 + 14 * 1 + 16 * 3 = 63 g / mol;
M (Ca (NO3) 2) = Mr (Ca (NO3) 2) = Ar (Ca) * N (Ca) + Ar (N) * N (N) + Ar (O) * N (O) = 40 * 1 + 14 * 2 + 16 * 3 * 2 = 164 g / mol;
3) n (CaO) = m (CaO) / M (CaO) = 14/56 = 0.25 mol;
4) n (HNO3) = m (HNO3) / M (HNO3) = 36/63 = 0.57 mol;
5) n (Ca (NO3) 2) = n (CaO) = 0.25 mol;
6) m (Ca (NO3) 2) = n (Ca (NO3) 2) * M (Ca (NO3) 2) = 0.25 * 164 = 41 g.

Answer: The mass of Ca (NO3) 2 is 41 g.