An excess of sodium hydroxide solution is added to 50 g of 8% nitric acid solution.

| September 25, 2021 | education

An excess of sodium hydroxide solution is added to 50 g of 8% nitric acid solution. Calculate the mass of the salt formed.

HNO3 + NaOH = NaNO3 + H2O.
Algorithm for solving the problem:
1) Calculation of the mass of the solute of nitric acid.
2) Calculation of the amount of nitric acid substance.
3) Finding the amount of sodium nitrate substance.
4) Calculation of the mass of sodium nitrate.
Solution:
1) The mass of the solute is found by multiplying the mass of the solution by the mass fraction of the substance: 50 * 8/100 = 4g.
2) The amount of substance is calculated by dividing the mass by the molar mass: 4/63 = 0.06 mol.
3) According to the reaction equation, there is one mole of sodium nitrate for one mole of nitric acid. This means that the amount of sodium nitrate substance is 0.06 mol.
4) Mass of sodium nitrate: 0.06 * 85 = 5.1 g – answer.



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