Question

for the following reaction, 67.1 grams of silver nitrate are allowed to react with 25.2 grams of copper(ii) chloride. silver nitrate (aq) + copper(ii) chloride (s) → silver chloride (s) + copper(ii) nitrate (aq) what is the maximum amount of silver chloride that can be formed? grams what is the formula for the limiting reagent? what amount of the excess reagent remains after the reaction is complete? grams

Explanation:

Step1: Balance the chemical equation

$2AgNO_3(aq)+CuCl_2(s)
ightarrow 2AgCl(s)+Cu(NO_3)_2(aq)$

Step2: Calculate the molar - masses

The molar mass of $AgNO_3$ is $M_{AgNO_3}=107.87 + 14.01+3\times16.00=169.88\ g/mol$.
The molar mass of $CuCl_2$ is $M_{CuCl_2}=63.55 + 2\times35.45 = 134.45\ g/mol$.
The molar mass of $AgCl$ is $M_{AgCl}=107.87+35.45 = 143.32\ g/mol$.

Step3: Calculate the number of moles of reactants

The number of moles of $AgNO_3$, $n_{AgNO_3}=\frac{m_{AgNO_3}}{M_{AgNO_3}}=\frac{67.1\ g}{169.88\ g/mol}\approx0.395\ mol$.
The number of moles of $CuCl_2$, $n_{CuCl_2}=\frac{m_{CuCl_2}}{M_{CuCl_2}}=\frac{25.2\ g}{134.45\ g/mol}\approx0.187\ mol$.

Step4: Determine the limiting reagent

From the balanced equation, the mole - ratio of $AgNO_3$ to $CuCl_2$ is $2:1$.
For $0.187\ mol$ of $CuCl_2$, the moles of $AgNO_3$ required is $2\times0.187\ mol = 0.374\ mol$. Since $0.395\ mol$ of $AgNO_3$ is present and only $0.374\ mol$ is required to react with $0.187\ mol$ of $CuCl_2$, $CuCl_2$ is the limiting reagent.

Step5: Calculate the moles of $AgCl$ formed

From the balanced equation, the mole - ratio of $CuCl_2$ to $AgCl$ is $1:2$.
Since $n_{CuCl_2}=0.187\ mol$, the moles of $AgCl$ formed, $n_{AgCl}=2\times n_{CuCl_2}=2\times0.187\ mol = 0.374\ mol$.
The mass of $AgCl$ formed, $m_{AgCl}=n_{AgCl}\times M_{AgCl}=0.374\ mol\times143.32\ g/mol\approx53.6\ g$.

Step6: Calculate the amount of excess reagent remaining

The moles of $AgNO_3$ that react is $0.374\ mol$.
The moles of $AgNO_3$ remaining, $n_{AgNO_3\ remaining}=0.395\ mol - 0.374\ mol=0.021\ mol$.
The mass of $AgNO_3$ remaining, $m_{AgNO_3\ remaining}=n_{AgNO_3\ remaining}\times M_{AgNO_3}=0.021\ mol\times169.88\ g/mol\approx3.67\ g$.

Answer:

What is the maximum amount of silver chloride that can be formed? 53.6 grams
What is the FORMULA for the limiting reagent? $CuCl_2$
What amount of the excess reagent remains after the reaction is complete? 3.67 grams