Question

When 22.4 L of $H_{2(g)}$ is mixed with 11.2 L of $Cl_{2(g)}$, each at STP, the moles of $HCl_{(g)}$ formed is equal to

• A. 1 mole of $HCl_{(g)}$
• B. 2 moles of $HCl_{(g)}$
• C. 0.5 mole of $HCl_{(g)}$
• D. 1.5 moles of $HCl_{(g)}$

Answer 1

Answer: (A)

1 mole of $HCl_{(g)}$

Answer 2

The given problem is related to the concept of
stoichiometry of chemical equations. Thus, we have
to convert the given volumes into their moles and
then, identify the limiting reagent [possessing
minimum number of moles and gets completely
used up in the reaction]. The limiting reagent gives
the moles of product formed in the reaction.
$H_2 (g) + Cl_2(g)\rightarrow 2HCl (g)$
Initial vol. $22.4 L\, \, 11.2 L\, \, 2 mol$
$\therefore \, \, \,$ 22.4 L volume at STP is occupied by
$\, \, \, \, \, \, \, \, \, \, \, Cl_2 = 1 \, mole$
$\therefore$ 11.2 L volume will be occupied by
$Cl_2 = \frac {1 \times 11.2}{22.4}mol \, = 0.5 \, mol$
22.4 L volume at STP is occupied by $H_2 = 1 \, mol$
Thus, $H_2 (g) + Cl_2 (g) \rightarrow 2HCl (g)$
$1 mol / / / 0.5 mol$
Since, $Cl_2$ possesses minimum number of moles,
thus it is the limiting reagent.
As per equation,
1 mole of $Cl_2 = 2$ moles of HC1
$\therefore$ 0.5 mole of $Cl_2 = 2 \times 0.5$ mole of HC1
$\, \, \, \, \, \, \, \, \, \, = 1.0 \, mole\, of \, HCl$
Hence, 1.0 mole of HC1 (g) is produced by
0.5 mole of $Cl_2$ [or 11.2 L].