Answer
A total of 1.67g of $Al$ was not consumed.
Work Step by Step
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 $Al$ is 3 to 2:
$0.0571 mol (Cl_2) \times \frac{ 2 mol(Al)}{ 3 mol (Cl_2)} = 0.0381mol (Al)$
2. Calculate the mass of $Al$:
26.98* 1 = 26.98g/mol
$0.0381 mol \times \frac{ 26.98 g}{ 1 mol} = 1.03g (Al)$
3. The initial $Al$ mass was 2.70g, and 1.03g was consumed:
2.70g - 1.03g = 1.67g
