Question

1. how many grams of cl₂ are there in 65.2 grams fe?

Explanation:

Step1: Find moles of Fe

$n_{Fe}=\frac{m_{Fe}}{M_{Fe}}=\frac{65.2\ g}{55.85\ g/mol}\approx1.17\ mol$

Step2: Determine mole - ratio

The reaction between Fe and $Cl_2$ is $2Fe + 3Cl_2=2FeCl_3$, mole - ratio $n_{Fe}:n_{Cl_2}=2:3$, so $n_{Cl_2}=\frac{3}{2}n_{Fe}$.
$n_{Cl_2}=\frac{3}{2}\times1.17\ mol = 1.755\ mol$

Step3: Calculate mass of $Cl_2$

$m_{Cl_2}=n_{Cl_2}\times M_{Cl_2}=1.755\ mol\times70.9\ g/mol\approx124\ g$

Answer:

124 g