Magnesium oxide in an amount of 0.1 mol was acted upon by 15 g of nitric acid. calculate the mass of the resulting salt.

Let’s find the amount of substance HNO3 by the formula:

n = m: M.

M (HNO3) = 63 g / mol.

n = 15 g: 63 g / mol = 0.23 mol.

HNO3 is given in excess. We solve for lack of (MgO).

Let’s compose the reaction equation, find the quantitative ratios of substances.

MgO + 2HNO3 = Mg (NO3) 2 + H2O.

For 1 mole of MgO, there is 1 mole of Mg (NO3) 2. The substances are in quantitative ratios of 1: 1.

n (MgO) = n (Mg (NO3) 2) = 0.1 mol

Let’s find the mass of Mg (NO3) 2:

m = n × M,

M (Mg (NO3) 2) = 148 g / mol.

m = 0.1 mol × 148 g / mol = 14.8 g.

Answer: 14.8 g.