Question

For each of the following balanced chemical equations, calculate how many grams of the products listed would be produced by complete reaction of 0.320 mol of the first reactant.

(a) AgNO₃(aq) + LiOH(aq) → AgOH(s) + LiNO₃(aq)

AgOH	g
LiNO3	g

Answer 1

(a)

AgNO₃(aq) + LiOH(aq) → AgOH(s) + LiNO₃(aq)

In this balanced chemical equations

1 mole of AgNO₃ reacts with 1 mole of LiOH to produce 1 mole of AgOH and 1 mole of  LiNO₃.

1 mole of AgNO₃ produces 1 mole of AgOH and 1 mole of  LiNO₃.

0.32 mole of AgNO₃ produces 0.32 mole of AgOH and 0.32 mole of  LiNO₃.

Molar mass of AgOH = 124.88 g/mol

1 mole of AgOH = 124.88 g

0.32 mole of AgOH = (0.32 x 124.88) g = 39.96 g

Molar mass of LiNO₃ = 68.95 g/mol

1 mole of  LiNO₃ = 68.95 g

0.32 mole of LiNO₃ = (0.32 x 68.95) g = 22.06 g

AgOH​ = 39.96 g and LiNO₃ = 22.06 g