Answer to Question #260334 in General Chemistry for kassy

Question

A solution containing 19.8 grams of barium chloride is mixed with a solution containing 22.4 grams of aluminum sulfate. What mass of barium sulfate will form?Which substance in the excess reagent? 3BaCl2 (aq) + Al2(SO4)3 (aq) → 3BaSO4 (s) + 2AlCl3 (aq *

Answer 1

Mass of aluminum sulfate = 115 g

Molar mass of aluminum sulfate = 342.15 g/mol

Moles of aluminum sulfate = 115 g / 342.15 g/mol = 0.336 mol

Mass of barium chloride = 225 g

Molar mass of barium chloride = 208.23 g/mol

Moles of barium chloride = 225 g / 208.23 g/mol = 1.08 mol

For complete reaction, the required mol ratio of aluminum sulfate : barium chloride = 1 : 3

Here, the available mol ratio of aluminum sulfate : barium chloride = 0.336 : 1.08

= 1 : 3.2

Thus, the available mol of barium chloride is more than the required mol of barium chloride.

Therefore, barium chloride is the excess reactant in the reaction.

This implies that aluminum sulfate is the limiting reactant in the reaction.

Mass of silver nitrate = 1.699 g

Molar mass of silver nitrate = 169.87 g/mol

Moles of silver nitrate = 1.699 g / 169.87 g/mol = 0.01000 mol

Theoretically, 3 mol of silver nitrate forms 3 mol of silver chloride

Therefore, 0.01000 mol of silver nitrate forms 0.01000 mol of silver chloride

Molar mass of silver chloride = 143.32 g/mol

Hence, the theoretical yield of silver chloride = 0.01000 mol x 143.32 g/mol = 1.433 g