3. five grams of copper metal react with a solution containing twenty grams of silver nitrate to produce…

Explanation:

Step1: Write the balanced chemical equation

$Cu + 2AgNO_3\rightarrow Cu(NO_3)_2+2Ag$

Step2: Calculate the molar - masses

The molar mass of $Cu$ is $M_{Cu}=63.55\ g/mol$, and the molar mass of $AgNO_3$ is $M_{AgNO_3}=107.87 + 14.01+3\times16.00=169.88\ g/mol$.

Step3: Calculate the number of moles of each reactant

The number of moles of $Cu$, $n_{Cu}=\frac{m_{Cu}}{M_{Cu}}=\frac{5\ g}{63.55\ g/mol}\approx0.0787\ mol$. The number of moles of $AgNO_3$, $n_{AgNO_3}=\frac{m_{AgNO_3}}{M_{AgNO_3}}=\frac{20\ g}{169.88\ g/mol}\approx0.118\ mol$.

Step4: Determine the limiting reactant

From the balanced equation, the mole - ratio of $Cu$ to $AgNO_3$ is $1:2$. For $0.0787\ mol$ of $Cu$, the moles of $AgNO_3$ required is $n_{AgNO_3\ required}=2\times n_{Cu}=2\times0.0787\ mol = 0.1574\ mol$. But we have only $0.118\ mol$ of $AgNO_3$. For $0.118\ mol$ of $AgNO_3$, the moles of $Cu$ required is $n_{Cu\ required}=\frac{n_{AgNO_3}}{2}=\frac{0.118\ mol}{2}=0.059\ mol$. Since we have $0.0787\ mol$ of $Cu$, $AgNO_3$ is the limiting reactant.

Step5: Calculate the amount of limiting reactant needed to react with excess reactant

We have $0.0787\ mol$ of $Cu$ (excess reactant). The moles of $AgNO_3$ needed to react completely with $0.0787\ mol$ of $Cu$ is $n = 2\times0.0787\ mol=0.1574\ mol$. The mass of $AgNO_3$ needed, $m = n\times M_{AgNO_3}=0.1574\ mol\times169.88\ g/mol\approx26.74\ g$.

Answer:

a. $AgNO_3$ b. $26.74\ g$