Question

2.0 g al and 2.0 g cl2
express your answer using two significant figures.
m =

submit request answer

part b
7.7 g al and 25.0 g cl2
express your answer using three significant figures.
m =

Explanation:

Step1: Write the chemical reaction

$2Al + 3Cl_{2}
ightarrow2AlCl_{3}$

Step2: Calculate molar - masses

The molar mass of $Al$ is $M_{Al}=26.98\ g/mol$, and the molar mass of $Cl_{2}$ is $M_{Cl_{2}} = 2\times35.45=70.90\ g/mol$, and the molar mass of $AlCl_{3}$ is $M_{AlCl_{3}}=26.98 + 3\times35.45=133.33\ g/mol$.

Step3: Calculate moles of reactants for Part A

For $Al$: $n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{2.0\ g}{26.98\ g/mol}\approx0.074\ mol$
For $Cl_{2}$: $n_{Cl_{2}}=\frac{m_{Cl_{2}}}{M_{Cl_{2}}}=\frac{2.0\ g}{70.90\ g/mol}\approx0.028\ mol$

Step4: Determine the limiting reactant for Part A

From the balanced equation, the mole - ratio of $Al$ to $Cl_{2}$ is $\frac{n_{Al}}{n_{Cl_{2}}}=\frac{2}{3}$.
The required moles of $Al$ for $0.028\ mol$ of $Cl_{2}$ is $n_{Al_{required}}=\frac{2}{3}n_{Cl_{2}}=\frac{2}{3}\times0.028\ mol\approx0.019\ mol$. Since $0.074\ mol>0.019\ mol$, $Cl_{2}$ is the limiting reactant.

Step5: Calculate moles of $AlCl_{3}$ formed for Part A

From the balanced equation, $n_{AlCl_{3}}=\frac{2}{3}n_{Cl_{2}}$. Substituting $n_{Cl_{2}} = 0.028\ mol$, we get $n_{AlCl_{3}}=\frac{2}{3}\times0.028\ mol\approx0.019\ mol$.

Step6: Calculate mass of $AlCl_{3}$ formed for Part A

$m_{AlCl_{3}}=n_{AlCl_{3}}\times M_{AlCl_{3}}=0.019\ mol\times133.33\ g/mol\approx2.5\ g$

Step7: Calculate moles of reactants for Part B

For $Al$: $n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{7.7\ g}{26.98\ g/mol}\approx0.285\ mol$
For $Cl_{2}$: $n_{Cl_{2}}=\frac{m_{Cl_{2}}}{M_{Cl_{2}}}=\frac{25.0\ g}{70.90\ g/mol}\approx0.353\ mol$

Step8: Determine the limiting reactant for Part B

The required moles of $Al$ for $0.353\ mol$ of $Cl_{2}$ is $n_{Al_{required}}=\frac{2}{3}n_{Cl_{2}}=\frac{2}{3}\times0.353\ mol\approx0.235\ mol$. Since $0.285\ mol>0.235\ mol$, $Cl_{2}$ is the limiting reactant.

Step9: Calculate moles of $AlCl_{3}$ formed for Part B

$n_{AlCl_{3}}=\frac{2}{3}n_{Cl_{2}}=\frac{2}{3}\times0.353\ mol\approx0.235\ mol$

Step10: Calculate mass of $AlCl_{3}$ formed for Part B

$m_{AlCl_{3}}=n_{AlCl_{3}}\times M_{AlCl_{3}}=0.235\ mol\times133.33\ g/mol\approx31.3\ g$

Answer:

Part A: $m = 2.5\ g$
Part B: $m = 31.3\ g$