Question

Consider the reaction: Mg (s) + AgNO3 (aq) ® Ag (s) + Mg(NO3)2 (aq). If you begin the reaction with 0.50 grams Mg and 19 mL of 0.025 M AgNO3, what will the theoretical yield of silver be?

Answer 1

Mg _(s) + 2 AgNO_{3 (aq)} $\to$ 2 Ag _(s) + Mg(NO₃)_{2 (aq)}

n(Mg) = 0.50 / 24.3 = 0.0206 mol

n(AgNO₃) = 19(0.025 / 1000) = 0.000475 mol

Mg is in excess

m(Ag) = n(AgNO₃) × M(Ag) = 0.000475 × 107.9 = 0.051 g

Answer: 0.051 g.