What mass of barium chloride and sodium sulfate should the solutions contain in order to get 23.3 g

What mass of barium chloride and sodium sulfate should the solutions contain in order to get 23.3 g of sediment when they are crossed?

First, we write down the reaction equation.
ВаСl2 + Na2SO4 -> 2NaCl + BaSO4
Let us determine the mass of barium chloride by the reaction equation.
M (BaCl2) = 208.246 g / mol.
M (BaSO4) = 233.43 g / mol.
X g BaCl2 – 23.3 g BaSO4
208.246 g / mol BaCl2 – 233.43 g / mol BaSO4
x = 208.246 × 23.3 ÷ 233.43 = 20.7 g BaCl2
Find the mass of sodium sulfate.
M (Na2SO4) = 142.04 g / mol.
X g Na2SO4 – 23.3 g BaSO4
142.04 g / mol Na2SO4 – 233.43 g / mol BaSO4
X = 142.04 × 23.3 ÷ 233.43 = 14.2 g Na2SO4
Answer: 14.2 g sodium sulfate and 20.7 g barium chloride.