Question

Question #e1f0a

Answer 1

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

There are 0.1357 mol of AgNO₃ in the solution.

The volume of NaCl required is 271 mL.

23.05 g AgNO₃ × `(1 mol AgNO₃)/(169.9 g AgNO₃)` = 0.1357 mol AgNO₃

NaCl(s) → Na⁺(aq) + Cl⁻(aq)

AgNO₃(s) → Ag⁺(aq) + NO₃⁻(aq)

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

0.1357 mol AgNO₃ × #(1 mol Ag⁺)/(1 mol AgNO₃) × (1 mol Cl⁻)/(1 mol Ag⁺)# = 0.1357 mol Cl⁻

0.1357 mol Cl⁻ × `(1 mol NaCl)/(1 mol Cl⁻) × ( 1 L NaCl)/(0.500 mol NaCl)` = 0.271 L NaCl =
271 mL NaCl