Question

for the following reaction, 4.67 grams of bromine are mixed with excess chlorine gas. the reaction yields 5.37 grams of bromine monochloride. bromine (g) + chlorine (g) → bromine monochloride (g) what is the theoretical yield of bromine monochloride? what is the percent yield for this reaction? 9 more group attempts remaining

Explanation:

Step1: Write the balanced chemical equation

The reaction is $Br_2(g)+Cl_2(g)
ightarrow 2BrCl(g)$. The molar - mass of $Br_2$ is $M_{Br_2}=2\times79.904\ g/mol = 159.808\ g/mol$, and the molar - mass of $BrCl$ is $M_{BrCl}=79.904 + 35.453=115.357\ g/mol$.

Step2: Calculate the moles of $Br_2$

The moles of $Br_2$ initially, $n_{Br_2}=\frac{m_{Br_2}}{M_{Br_2}}=\frac{4.67\ g}{159.808\ g/mol}=0.0292\ mol$.

Step3: Calculate the theoretical yield of $BrCl$

From the balanced equation, 1 mole of $Br_2$ produces 2 moles of $BrCl$. So, the moles of $BrCl$ produced theoretically, $n_{BrCl}^{theo}=2\times n_{Br_2}=2\times0.0292\ mol = 0.0584\ mol$. The theoretical mass of $BrCl$, $m_{BrCl}^{theo}=n_{BrCl}^{theo}\times M_{BrCl}=0.0584\ mol\times115.357\ g/mol = 6.74\ g$.

Step4: Calculate the percent yield

The percent yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$. The actual yield of $BrCl$ is $5.37\ g$. So, $\text{Percent Yield}=\frac{5.37\ g}{6.74\ g}\times100\% = 79.7\%$.

Answer:

The theoretical yield of bromine monochloride is $6.74$ grams and the percent yield is $79.7\%$.