how would i solve this?

Question

Barium sulphide will be prepared by reduction of barium sulphate by carbon. How many grams of barium sulphate would you weigh to prepare 25 g of barium sulphide?

Answer 1

The balanced equation for this reaction is

BaSO4 + 2 C → BaS + 2CO2

25 g BaS x 1 mole/169 g = 0.148 moles

moles BaSO4 needed = 0.148 mol BaS x1 mol BaSO4/moleBaS = 0.148 moles

grams BaSO4 needed = 0.148 moles x 233g/mole = 34.5 g = 35 g (to 2 sig. figs.)

Answer 2

Hi Kado, to solve, write a balanced chemical equation and use the stoichiometry to determine how many grams of barium sulphate are necessary to yield 25 g of barium sulphide.