Question

1. consider the following reaction to answer the questions below: solid aluminum reacts with aqueous copper (ii) chloride to produce aqueous aluminum chloride and solid copper.

a. write and balance the equation to represent the reaction, with all appropriate notation
b. classify the reaction
c. identify the limiting reactant if you start with 7.17 g of aluminum and 68.9 g of copper chloride
d. determine how many moles of the excess reactant will remain unused
e. how many moles of each product are formed?
f. what is the mass of each product formed?
g. if this reaction is conducted in lab and 21.5 g of copper are produced, what is the percent yield?

Explanation:

Step1: Write and balance the equation

The un - balanced equation is $Al(s)+CuCl_2(aq)
ightarrow AlCl_3(aq)+Cu(s)$.
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 $AlCl_3$: $Al(s)+3CuCl_2(aq)
ightarrow 2AlCl_3(aq)+Cu(s)$. 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 $2Al(s)+3CuCl_2(aq)
ightarrow 2AlCl_3(aq)+3Cu(s)$.

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 $2Al(s)+3CuCl_2(aq)
ightarrow 2AlCl_3(aq)+3Cu(s)$, 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. $2Al(s)+3CuCl_2(aq)
ightarrow 2AlCl_3(aq)+3Cu(s)$
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. $84.8\%$