A) \[89.6\] L
B) \[67.2\] L
C) \[44.8\] L
D) \[22.4\] L
Correct Answer: B
Solution :
\[\underset{2\,\,mol}{\mathop{2BC{{l}_{3}}}}\,+\underset{3\,\,mol}{\mathop{3{{H}_{2}}}}\,\xrightarrow{{}}\underset{2\,\,mol}{\mathop{2B}}\,+6HCl\] 21.6 g = 2 mol 21.6 g of B = 2 mol of B \[\equiv \] 3 mol of \[{{H}_{2}}\] pV = nRT \[\therefore \] \[V=\frac{nRT}{p}=\frac{3\times 0.0821\times 273}{1}=67.2\,L\]You need to login to perform this action.
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