Determine the mass of the salt formed by the interaction of 45 g of nitric acid solution

Determine the mass of the salt formed by the interaction of 45 g of nitric acid solution with calcium hydroxide. (mass fraction of acid in solution 73%)

Let’s implement the solution:
According to the condition of the problem, we compose the equation of the process:
2HNO3 + Ca (OH) 2 = Ca (NO3) 2 + 2H2O – ion exchange, calcium nitrate was formed;
Calculations:
M (HNO3) = 63 g / mol;
M Ca (NO3) 2 = 164 g / mol.
Let’s determine the mass, the amount of acid, taking into account the product yield:
m (HNO3) = 45 / 0.73 = 61.6 g;
Y (HNO3) = m / M = 61.6 / 63 = 0.97 mol.
Proportion:
0.97 mol (HNO3) – X mol Ca (NO3) 2;
-2 mol                     – 1 mol hence, X mol Ca (NO3) 2 = 0.97 * 1/2 = 0.49 mol.
Find the mass of salt:
m Ca (NO3) 2 = Y * M = 0.49 * 164 = 80.36 g
Answer: received calcium nitrate weighing 80.36 g