Question

Find the moles of HCl gas obtained at STP from the following set of reactions:
$FeC{l}_3+O_2\to F{e}_2{O}_3+C{l}_2\left(g\right)H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$
Starting with 4.56 mol of $FeC{l}_3$ , 33.6 L of $O_2$ at STP and 7 mol of $H_2$ gas.
(Molar mass of Fe = 56 g/mol)

• A. 2 mol
• B. 4 mol
• C. 6 mol
• D. 8 mol

Answer 1

The correct option is C 6 mol

Balancing the given equation:
$2FeC{l}_3+\frac{3}{2}{O}_2\to F{e}_2{O}_3+3C{l}_2\left(g\right)H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$
$Molesof{O}_2=\frac{\text{given volume at STP}}{\text{molar volume at STP}}=\frac{33.6}{22.4}=1.5mol$

Finding limiting reagent:
$ForFeC{l}_3=\frac{\text{Given moles}}{\text{stoichiometric coefficient}}=\frac{4.56}{2}=2.28\text{For}{O}_2=\frac{\text{given moles}}{\text{stoichiometric coefficient}}=\frac{1.5}{1.5}=1.0$
So, oxygen will be the limiting reagent.
According to stoichiometry,
1.5 mol of $O_2$ produces 3 mol of $C{l}_2\text{gas.}$
In reaction (ii),
Finding the limiting reagent:
$For{H}_2=\frac{\text{Given moles}}{\text{stoichiometric coefficient}}=\frac{7}{1}=7mol\text{For}C{l}_2=\frac{\text{Given moles}}{\text{stoichiometri coefficient}}=\frac{3}{1}=3mol$
So, chlorine will be the limiting reagent.
1 mol of $C{l}_2$ produces 2 mol of $HCl$.
3 mol will produce 6 mol of $HCl$.