Question

for the following reaction, 31.1 grams of bromine are allowed to react with 12.1 grams of chlorine gas. bromine (g) + chlorine (g) → bromine monochloride (g) what is the maximum mass of bromine monochloride that can be formed? mass = g what is the formula for the limiting reactant? what mass of the excess reactant remains after the reaction is complete? mass = g

Explanation:

Step1: Write the balanced chemical equation

$Br_2(g)+Cl_2(g)
ightarrow 2BrCl(g)$

Step2: Calculate the moles of reactants

The molar - mass of $Br_2$ is $M_{Br_2}=2\times79.904\ g/mol = 159.808\ g/mol$, and the moles of $Br_2$ is $n_{Br_2}=\frac{31.1\ g}{159.808\ g/mol}\approx0.195\ mol$. The molar - mass of $Cl_2$ is $M_{Cl_2}=2\times35.45\ g/mol = 70.9\ g/mol$, and the moles of $Cl_2$ is $n_{Cl_2}=\frac{12.1\ g}{70.9\ g/mol}\approx0.171\ mol$.

Step3: Determine the limiting reactant

From the balanced equation, the mole ratio of $Br_2$ to $Cl_2$ is $1:1$. Since $n_{Cl_2}

Step4: Calculate the mass of $BrCl$ formed

The molar - mass of $BrCl$ is $M_{BrCl}=79.904 + 35.45=115.354\ g/mol$. From the balanced equation, the mole ratio of $Cl_2$ to $BrCl$ is $1:2$. The moles of $BrCl$ formed is $n_{BrCl}=2\times n_{Cl_2}=2\times0.171\ mol = 0.342\ mol$. The mass of $BrCl$ formed is $m_{BrCl}=n_{BrCl}\times M_{BrCl}=0.342\ mol\times115.354\ g/mol\approx39.4\ g$.

Step5: Calculate the moles of excess reactant remaining

The moles of $Br_2$ that react with $0.171\ mol$ of $Cl_2$ is $0.171\ mol$. The moles of $Br_2$ remaining is $n_{Br_2,remaining}=n_{Br_2}-n_{Br_2,reacted}=0.195\ mol - 0.171\ mol = 0.024\ mol$. The mass of $Br_2$ remaining is $m_{Br_2,remaining}=n_{Br_2,remaining}\times M_{Br_2}=0.024\ mol\times159.808\ g/mol\approx3.84\ g$.

Answer:

The formula for the limiting reactant is $Cl_2$. The maximum mass of bromine monochloride that can be formed is approximately $39.4\ g$. The mass of the excess reactant ($Br_2$) remaining after the reaction is complete is approximately $3.84\ g$.