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Question
When 22.4 litres of H2 (g) is mixed with 11.2 litres of Cl2 (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to
Options
2 moles of HCl (g)
0.5 moles of HCl (g)
1.5 moles of HCl (g)
1 mole of HCl (g)
Solution
1 mole of HCl (g)
Explanation:
\[\ce{H2_{(g)} + Cl2_{(g)} -> 2HCl_{(g)}}\]
| Content | \[\ce{H2_{(g)}}\] | \[\ce{Cl2_{(g)}}\] | \[\ce{HCl_{(g)}}\] |
| Stoichiometric coefficient | 1 | 1 | 2 |
| No. of moles of reactants allowed to react at 273 K and 1 atm pressure | 22.4 L (1 mol) |
11.2 L (0.5 mol) |
- |
| No. of moles of reactant reacted and product formed | 0.5 | 0.5 | 1 |
| Amount of HCl formed = 1 mol | |||
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