A) 1 mole of \[HCl(g)\]
B) 2 moles of \[HCl(g)\]
C) 0.5 mole of \[HCl(g)\]
D) 1.5 moleof \[HCl(g)\]
Correct Answer: B
Solution :
| 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. |
| \[\underset{\text{lnitial}\,\text{vol}\text{.}}{\overset{{}}{\mathop{{}}}}\,\underset{\text{22}\text{.4L}}{\mathop{\,\,\,{{\text{H}}_{\text{2}}}\text{(g)}}}\,\text{+}\underset{\text{11}\text{.2L}}{\mathop{\text{C}{{\text{l}}_{\text{2}}}\text{(g)}}}\,\to \underset{\text{2mol}}{\mathop{\text{2HCl(g)}}}\,\] |
| \[\because \] 22.4 L volume at STP is occupied by |
| \[C{{l}_{2}}\]= 1 mole, |
| \[\therefore \]11.2 L volume will be occupied by, |
| \[C{{l}_{2}}=\frac{1\times 11.2}{22.4}\text{mole}\] |
| = 0.5 mol |
| Thus, \[\underset{1\,\text{mol}}{\mathop{{{H}_{2}}(g)}}\,+\underset{0.5\text{mol}}{\mathop{C{{l}_{2}}(g)}}\,\to 2HCl(g)\] |
| Since, \[C{{l}_{2}}\] possesses minimum number of moles, thus it is the limiting reagent. |
| As per equation, |
| \[\text{1}\,\text{mol}\,\text{C}{{\text{l}}_{\text{2}}}\,\equiv \,\text{2}\,\text{mol}\,\text{HCl}\] |
| \[\therefore \] \[0.5molC{{l}_{2}}=2\times 0.5molHCl\] |
| \[=1.0\text{molHCl}\] |
| Hence, 1.0 mole of HCI (g) is produced by 0.5 mole of \[C{{l}_{2}}\] [or 11.2 L]. |
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