The slaked lime, taken in the required amount, was acted upon by 4 kg of pure nitric acid.

The slaked lime, taken in the required amount, was acted upon by 4 kg of pure nitric acid. What is the mass of calcium nitrate Ca (NO3) 2 obtained if the practical yield in mass fractions is 0.9 in comparison with the theoretical one?

The equation for the reaction of slaked lime with nitric acid:
Ca (OH) 2 + 2HNO3 = Ca (NO3) 2 + 2H2O.
Molar mass of nitric acid:
1 + 14 + 16 * 3 = 63 (g / mol).
Molar mass of slaked lime:
40+ (1 + 16) * 2 = 74 (g / mol).
From the reaction equation:
74 g Ca (OH) 2 – 126 g HNO3;
4000 g Ca (OH) 2 – x g HNO3;
x = 4000 * 126/74 = 6810.81 (g).
Let’s calculate a practical way out:
6810.81 * 0.9 = 6129.73 (g).
Answer: 6129.73 g.