Determine the mass of the salt obtained by the interaction of 60 g of HNO3 with 0.5 mol of NaOH.

To solve, we write down the equation of the process:

HNO3 + NaOH = NaNO3 + H2O – ion exchange, sodium nitrate was formed;
Let’s make the calculations:
M (HNO3) = 63 g / mol;

M (NaOH) = 39.9 g / mol;

M (NaNO3) = 84.9 g / mol.

Determine the amount of acid:
Y (HNO3) = m / M = 60/63 = 0.95 mol (substance in excess);

Y (NaOH) = 0.5 mol (deficient substance);

Y (NaNO3) = 0.5 mol since the amount of substances is 1 mol.

The calculation is based on the substance in deficiency.

4. Find the mass of the product:

m (NaNO3) = Y * M = 0.5 * 84.9 = 42.45 g

Answer: received sodium nitrate weighing 42.45 g