Question

How many grams of Na2SO4 are needed to precipitate all the barium ions produced by 43.9g of Ba(NO3)2​

Answer 1

Explanation:

This is a stoichiometry problem. The general procedure is to convert Ba(NO3)2 grams → Ba(NO3)2 moles → Na2SO4 moles → Na2SO4 grams. Assuming that this will be a complete reaction, the amount of grams of Na2SO4 will be the minimum amount needed to precipitate all of the barium ions.

The conversion for Ba(NO3)2 grams → Ba(NO3)2 moles requires the molar mass of Ba(NO3)2 which is 426.7 g/mol. Similarly, the conversion for Na2SO4 moles → Na2SO4 grams requires the molar mass of Na2SO4 which is 142.05 g/mol. To convert Ba(NO3)2 moles → Na2SO4 moles, you use a balanced chemical equation which you have provided in your question, 1 mol Na2SO4 : 1 mol Ba(NO3)2.

Using stoichiometry,

43.9 g Ba(NO3)2 × (1 mol Ba(NO3)2 / 426.7 g Ba(NO3)2) × (1 mol Na2SO4 / 1 mol Ba(NO3)2) × (142.05 g Na2SO4 / 1 mol Na2SO4) = 14.6 g Na2SO4