Answer
$5.08g (Al_2Cl_6)$
Work Step by Step
As we determined in the last exercise, $Cl_2$ is the limiting reactant, so we have to do the calculations based on its mass.
1. Calculate the number of moles of $Cl_2$:
35.45* 2 = 70.9g/mol
$4.05g \times \frac{1 mol}{ 70.9g} = 0.0571mol (Cl_2)$
- The balanced reaction is:
$2Al + 3Cl_2 -- \gt Al_2Cl_6$
The ratio of $Cl_2$ to $Al_2Cl_6$ is 3 to 1:
$0.0571 mol (Cl_2) \times \frac{ 1 mol(Al_2Cl_6)}{ 3 mol (Cl_2)} = 0.0190mol (Al_2Cl_6)$
2. Calculate the mass of $Al_2Cl_6$:
26.98* 2 + 35.45* 6 = 266.66g/mol
$0.0190 mol \times \frac{ 266.66 g}{ 1 mol} = 5.08g (Al_2Cl_6)$
