A solution containing 63 g of nitric acid was added to a solution containing 85.5 g of barium hydroxyl.

| July 25, 2021 | education

A solution containing 63 g of nitric acid was added to a solution containing 85.5 g of barium hydroxyl. Determine the mass of the salt formed.

1. The interaction of barium hydroxide with nitric acid leads to the formation of barium nitrate:

Ba (OH) 2 + 2HNO3 = Ba (NO3) 2 + 2H2O;

2.Calculate the chemical quantities of reactants:

n (Ba (OH) 2) = m (Ba (OH) 2): M (Ba (OH) 2);

M (Ba (OH) 2) = 137 + 2 * 17 = 171 g / mol;

n (Ba (OH) 2) = 85.5: 171 = 0.5 mol;

n (HNO3) = m (HNO3): M (HNO3);

M (HNO3) = 1 + 14 + 3 * 16 = 63 g / mol;

n (HNO3) = 63: 63 = 1 mol;

3. Determine the amount of salt formed:

n (Ba (NO3) 2) = n (Ba (OH) 2) = 0.5 mol;

4. find the mass of the resulting barium nitrate:

m (Ba (NO3) 2) = n (Ba (NO3) 2) * M (Ba (NO3) 2);

M (Ba (NO3) 2) = 137 + 2 * 14 + 3 * 2 * 16 = 261 g / mol;

m (Ba (NO3) 2) = 0.5 * 261 = 130.5 g.

Answer: 130.5 g.



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