Question

0.230 g al and 1.20 g cl₂. express your answer using three significant figures. m = g

Explanation:

Step1: Write the chemical reaction

$2Al + 3Cl_2
ightarrow2AlCl_3$

Step2: Calculate the molar - mass of Al and $Cl_2$

The molar - mass of Al, $M_{Al}=26.98\ g/mol$, and the molar - mass of $Cl_2$, $M_{Cl_2}=2\times35.45 = 70.90\ g/mol$.

Step3: Calculate the number of moles of Al and $Cl_2$

The number of moles of Al, $n_{Al}=\frac{m_{Al}}{M_{Al}}=\frac{0.230\ g}{26.98\ g/mol}\approx0.008525\ mol$.
The number of moles of $Cl_2$, $n_{Cl_2}=\frac{m_{Cl_2}}{M_{Cl_2}}=\frac{1.20\ g}{70.90\ g/mol}\approx0.01693\ mol$.

Step4: Determine the limiting reactant

From the balanced chemical equation, the mole ratio of Al to $Cl_2$ is $\frac{n_{Al}}{n_{Cl_2}}=\frac{2}{3}$.
If all 0.008525 mol of Al reacts, the moles of $Cl_2$ required, $n_{Cl_2\ required}=\frac{3}{2}n_{Al}=\frac{3}{2}\times0.008525\ mol = 0.01279\ mol$. Since $0.01279\ mol<0.01693\ mol$, Al is the limiting reactant.

Step5: Calculate the moles of $AlCl_3$ formed

From the balanced chemical equation, the mole ratio of Al to $AlCl_3$ is 1:1. So, the moles of $AlCl_3$ formed, $n_{AlCl_3}=n_{Al}=0.008525\ mol$.

Step6: Calculate the molar - mass of $AlCl_3$

The molar - mass of $AlCl_3$, $M_{AlCl_3}=26.98\ g/mol+3\times35.45\ g/mol=26.98\ g/mol + 106.35\ g/mol = 133.33\ g/mol$.

Step7: Calculate the mass of $AlCl_3$ formed

$m_{AlCl_3}=n_{AlCl_3}\times M_{AlCl_3}=0.008525\ mol\times133.33\ g/mol\approx1.14\ g$.

Answer:

$1.14$