Mg (s) + 2 AgNO3 (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.
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