200 g of 10% solution of nitric acid NaOH how many grams of salt as a result of the interaction.

HNO3 + NaOH = NaNO3 + H2O.

The mass of nitric acid (pure) in solution is 10% of the total mass:

m (HNO3) = 200 * 0.1 = 20 g.

The amount of nitric acid:

n (HNO3) = m (HNO3) / M (HNO3).

n (HNO3) = 20/63 = 0.3174 mol.

Accordingly, based on the stoichiometric coefficients in the reaction equation, we see that the substances interact with each other in a ratio of 1 to 1. Then the amount of salt (sodium nitrate) will be:

n (NaNO3) = n (HNO3) = 0.3174 mol.

m (NaNO3) = m (NaNO3) * M (NaNO3) = 84.9947 * 0.3174 = 26.9773 g.