10 g of magnesium oxide was treated with a solution containing 40 g of nitric acid. How much salt is formed?

Given:
m (MgO) = 10 g
m (HNO3) = 40 g

To find:
m (salt) -?

Solution:
1) MgO + 2HNO3 => Mg (NO3) 2 + H2O;
2) M (MgO) = Mr (MgO) = Ar (Mg) + Ar (O) = 24 + 16 = 40 g / mol;
M (HNO3) = Mr (HNO3) = Ar (H) + Ar (N) + Ar (O) * 3 = 1 + 14 + 16 * 3 = 63 g / mol;
M (Mg (NO3) 2) = Mr (Mg (NO3) 2) = Ar (Mg) + Ar (N) * 2 + Ar (O) * 3 * 2 = 24 + 14 * 2 + 16 * 3 * 2 = 148 g / mol;
3) n (MgO) = m (MgO) / M (MgO) = 10/40 = 0.25 mol;
n (HNO3) = m (HNO3) / M (HNO3) = 40/63 = 0.63 mol;
4) n (Mg (NO3) 2) = n (MgO) = 0.25 mol;
5) m (Mg (NO3) 2) = n (Mg (NO3) 2) * M (Mg (NO3) 2) = 0.25 * 148 = 37 g.

Answer: The mass of Mg (NO3) 2 is 37 g.