2) 12.5 g of copper are reacted with an excess of chlorine gas, and 25.4 g of copper(ii) chloride are…

Explanation:

Step1: Identify limiting reactant

Chlorine is in excess, so copper (Cu) is the limiting reactant.

Step2: Calculate moles of Cu

The molar - mass of Cu is $M_{Cu}=63.55\ g/mol$. The number of moles of Cu, $n_{Cu}=\frac{m_{Cu}}{M_{Cu}}=\frac{12.5\ g}{63.55\ g/mol}\approx0.197\ mol$.

Step3: Determine moles of $CuCl_2$ from mole - ratio

From the balanced equation $Cu + Cl_2\rightarrow CuCl_2$, the mole - ratio of $Cu$ to $CuCl_2$ is 1:1. So, $n_{CuCl_2}=n_{Cu}\approx0.197\ mol$.

Step4: Calculate theoretical yield of $CuCl_2$

The molar - mass of $CuCl_2$ is $M_{CuCl_2}=63.55\ g/mol+(2\times35.45\ g/mol)=134.45\ g/mol$. The theoretical yield, $m_{theo}=n_{CuCl_2}\times M_{CuCl_2}=0.197\ mol\times134.45\ g/mol\approx26.5\ g$.

Step5: Calculate mass of excess reactant used

From the balanced equation, 1 mole of Cu reacts with 1 mole of $Cl_2$. Moles of $Cl_2$ used is equal to moles of Cu used, $n_{Cl_2}=0.197\ mol$. The molar - mass of $Cl_2$ is $M_{Cl_2}=2\times35.45\ g/mol = 70.9\ g/mol$. Mass of $Cl_2$ used, $m_{Cl_2}=n_{Cl_2}\times M_{Cl_2}=0.197\ mol\times70.9\ g/mol\approx13.9\ g$.

Step6: Calculate percent yield

The percent yield is given by the formula $\text{Percent Yield}=\frac{m_{actual}}{m_{theo}}\times100%$. Given $m_{actual} = 25.4\ g$ and $m_{theo}\approx26.5\ g$. $\text{Percent Yield}=\frac{25.4\ g}{26.5\ g}\times100%\approx95.8%$.

Answer:

a) $Cu + Cl_2\rightarrow CuCl_2$ b) Copper (Cu) c) $26.5\ g$ d) $13.9\ g$ e) $95.8%$