# What mass of magnesium nitrate is formed by the action of 8 grams of magnesium oxide with nitric

**What mass of magnesium nitrate is formed by the action of 8 grams of magnesium oxide with nitric acid, if the magnesium oxide contains 15% impurities?**

Let’s execute the solution:

1. According to the condition of the problem, we write the equation:

m = 8 g; 15% impurity. X g -?

MgO + 2HNO3 = Mg (NO3) 2 + H2O – exchange, magnesium nitrate was obtained;

2. Calculations:

M (MgO) = 40.3 g / mol.

M Mg (NO3) 2 = 148.3 g / mol.

3. Determine the mass of the original substance without impurity, its amount:

m (MgO) = 8 (1-0.15) = 6.8 g;

Y (MgO) = m / M = 6.8 / 40.3 = 0.17 mol;

Y Mg (NO3) 2 = 0.17 mol since the amount of substances according to the equation is 1 mol.

4. Find the mass of the product:

m Mg (NO3) 2 = Y * M = 0.17 * 148.3 = 25.2 g.

Answer: obtained magnesium nitrate weighing 25.2 g.