Question

A 115.0 −mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 230.0 −mL sample of a solution that is 0.10 M in NaCN.

Answer 1

The reaction:

AgNO₃ + NaCN -----------> AgCN + NaNO₃

Volume of AgNO₃ = 115.0 mL

Molarity of AgNO₃ = 3.0x10^-3 M

Volume of NaCN = 230.0 mL

Molarity of NaCN = 0.10 M

Number of moles of AgNO₃ = 3.0x10^-3 M x 115.0 mL = 0.345 mmoles

Number of moles of NaCN = 0.10M x 230.0 mL = 23 mmoles

Limiting reagent = AgNO₃

Excess Reagent = NaCN

After the completion of the reaction:

Amount of AgNO₃ remaining = Nil

Amount of NaCNremaining = 23-0.345 =22.655 mmoles

Amount of AgCN produced = 0.345 moles

Amount of NaNO₃ produced = 0.345 moles