5. consider the following reaction to answer the questions below: solid aluminum reacts with aqueous copper…

Explanation:

Step1: Write and balance the equation

The un - balanced equation is $Al(s)+CuCl_2(aq)\rightarrow AlCl_3(aq)+Cu(s)$. To balance the chlorine atoms, we find the least - common multiple of 2 and 3 (the sub - scripts of chlorine in $CuCl_2$ and $AlCl_3$), which is 6. So we put a 3 in front of $CuCl_2$ and a 2 in front of $AlCl_3$: $Al(s)+3CuCl_2(aq)\rightarrow 2AlCl_3(aq)+Cu(s)$. Then we balance the aluminum and copper atoms by putting a 2 in front of $Al$ and a 3 in front of $Cu$. The balanced equation is $2Al(s)+3CuCl_2(aq)\rightarrow 2AlCl_3(aq)+3Cu(s)$.

Step2: Classify the reaction

This is a single - replacement reaction. In a single - replacement reaction, one element replaces another element in a compound. Here, aluminum ($Al$) replaces copper ($Cu$) in copper(II) chloride ($CuCl_2$).

Step3: Calculate the moles of reactants

The molar mass of $Al$ is $M_{Al}=26.98\ g/mol$, and the molar mass of $CuCl_2$ is $M_{CuCl_2}=63.55 + 2\times35.45=134.45\ g/mol$. The number of moles of $Al$, $n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{7.17\ g}{26.98\ g/mol}=0.266\ mol$. The number of moles of $CuCl_2$, $n_{CuCl_2}=\frac{m_{CuCl_2}}{M_{CuCl_2}}=\frac{68.9\ g}{134.45\ g/mol}=0.512\ mol$.

Step4: Determine the limiting reactant

From the balanced equation $2Al(s)+3CuCl_2(aq)\rightarrow 2AlCl_3(aq)+3Cu(s)$, the mole ratio of $Al$ to $CuCl_2$ is $\frac{n_{Al}}{n_{CuCl_2}}=\frac{2}{3}$. If $Al$ is the limiting reactant, the moles of $CuCl_2$ required to react with $0.266\ mol$ of $Al$ is $n_{CuCl_2\ required}=0.266\ mol\times\frac{3}{2}=0.399\ mol$. Since $0.399\ mol<0.512\ mol$, $Al$ is the limiting reactant.

Step5: Calculate the moles of excess reactant remaining

The moles of $CuCl_2$ that react with $0.266\ mol$ of $Al$ is $n_{CuCl_2\ reacted}=0.266\ mol\times\frac{3}{2}=0.399\ mol$. The moles of $CuCl_2$ remaining, $n_{CuCl_2\ remaining}=n_{CuCl_2}-n_{CuCl_2\ reacted}=0.512\ mol - 0.399\ mol = 0.113\ mol$.

Step6: Calculate the moles of products formed

From the balanced equation, the mole ratio of $Al$ to $AlCl_3$ is 1:1, so the moles of $AlCl_3$ formed, $n_{AlCl_3}=0.266\ mol$. The mole ratio of $Al$ to $Cu$ is 2:3, so the moles of $Cu$ formed, $n_{Cu}=0.266\ mol\times\frac{3}{2}=0.399\ mol$.

Step7: Calculate the mass of products formed

The molar mass of $AlCl_3$ is $M_{AlCl_3}=26.98+3\times35.45 = 133.33\ g/mol$. The mass of $AlCl_3$ formed, $m_{AlCl_3}=n_{AlCl_3}\times M_{AlCl_3}=0.266\ mol\times133.33\ g/mol = 35.47\ g$. The molar mass of $Cu$ is $M_{Cu}=63.55\ g/mol$. The mass of $Cu$ formed, $m_{Cu}=n_{Cu}\times M_{Cu}=0.399\ mol\times63.55\ g/mol = 25.36\ g$.

Step8: Calculate the percent yield

The theoretical yield of $Cu$ is $25.36\ g$ and the actual yield is $21.5\ g$. The percent yield, $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100%=\frac{21.5\ g}{25.36\ g}\times100% = 84.8%$.

Answer:

a. $2Al(s)+3CuCl_2(aq)\rightarrow 2AlCl_3(aq)+3Cu(s)$ b. Single - replacement reaction c. Aluminum ($Al$) is the limiting reactant d. $0.113\ mol$ of $CuCl_2$ remains unused e. $0.266\ mol$ of $AlCl_3$ and $0.399\ mol$ of $Cu$ are formed f. $35.47\ g$ of $AlCl_3$ and $25.36\ g$ of $Cu$ are formed g. $84.8%$