Question

How much barium sulfate can be produced from the reaction of 10.0 g of sodium sulfate and an excess of barium chloride?

How much barium chloride was needed in the problem above?

Answer 1

How much barium sulfate can be produced from the reaction of 10.0 g of sodium sulfate and an excess of barium chloride?

How much barium chloride was needed in the problem above?

The recation of sodium sulfate and barium chloride is as follows:

Na2SO4 + BaCl2 → BaSO4 + 2 NaCl

Moles of Na2SO4 = 10.0 g molar mass

= 10.0 g / 142.04 g/mol

= 0.070 moles Na2SO4

Moles of BaSO4:

0.070 moles Na2SO4 * 1 moles BaSO4 / 1 mole Na2SO4

=0.070 moles BaSO4

Amount of BaSO4 = number of moles * molar mass

= 0.070 moles BaSO4*233.43 g/mol

= 16.43 g BaSO4

Moles of BaCl2:

0.070 moles Na2SO4 * 1 moles BaCl2 / 1 mole Na2SO4

=0.070 moles BaCl2

Amount of BaCl2 = number of moles * molar mass

= 0.070 moles BaCl2 *208.233 g/mol

= 14.57g BaCl2