Answer
$Cl_2$ is the limiting reactant.
Work Step by Step
1. Calculate the number of moles of $Al$:
26.98* 1 = 26.98g/mol
$2.70g \times \frac{1 mol}{ 26.98g} = 0.1mol (Al)$
- The balanced reaction is:
$2Al + 3O_2 -- \gt Al_2O_6$
The ratio of $Al$ to $Cl_2$ is 2 to 3:
$0.100 mol (Al) \times \frac{ 3 mol(Cl_2)}{ 2 mol (Al)} = 0.150mol (Cl_2)$
2. Calculate the mass of $Cl_2$:
35.45* 2 = 70.90g/mol
$0.150 mol \times \frac{ 70.90 g}{ 1 mol} = 10.6g (Cl_2)$
The amount of $Cl_2$ needed is higher than 4.05g.
Therefore, $Cl_2$ is the limiting reactant.
