How much barium nitrate can be obtained if 74 g reacts with nitric acid. barium containing 19% impurities?

Solution:

1. Let’s write the equation of the reaction of interaction of barium with nitric acid:

Ba + 2HNO3 = Ba (NO3) 2 + H2;

1. Let’s find the mass of pure barium:

74 * 0.81 = 59.94 grams;

1. Let’s find the amount of substance in barium:

59.94 / 137 = 0.44 mol;

1. Find the molecular weight of barium nitrate:

137 + 14 * 2 + 16 * 6 = 261 g / mol;

1. Before barium nitrate there is the same coefficient as before barium. This means that the amount of substance in barium nitrate is 0.44 mol. Let’s find the mass of barium nitrate:

261 * 0.44 = 114.84 grams;

Answer: The mass of barium nitrate is 114.84 grams.