Answer:
The answer to your question is 17.2 g of AgCl
Explanation:
Data
mass of NaCl = 7 g
mass of AgNO₃ = 95 g
mass of AgCl = ?
Balanced chemical reaction
NaCl + AgNO₃ ⇒ AgCl + NaNO₃
Process
1.- Calculate the molar mass of the reactants
NaCl = 23 + 35.5 = 58.5 g
AgNO₃ = 108 + 14 + 48 = 170 g
2.- Calculate the limiting reactant
theoretical yield = AgNO₃ / NaCl = 170 / 58.5 = 2.9
experimental yield = AgNO₃ / NaCl = 95 / 7 = 13.6 g
From these results we conclude that the limiting reactant is NaCl because the experimental yield was lower than the theoretical yield.
3.- Calculate the mass of AgCl
Molar mass of AgCl = 108 + 35.5 = 143.5 g
-Use proportions
58.5 g of NaCl ------------------ 143.5 g of AgCl
7.0 g of NaCl ------------------- x
x = (7 x 143.5) / 58.5
x = 17.2 g of AgCl
