A) 22.4 L
B) 89.6 L
C) 67.2 L
D) 44.8 L
Correct Answer: C
Solution :
\[BC{{l}_{3}}+3[H]\to B+3HCl\] \[BC{{l}_{3}}+\frac{3}{2}{{H}_{2}}\to B+3HCl\]; \[B=\frac{21.6}{10.8}=2\]mole \[B\equiv \frac{3}{2}{{H}_{2}}\] 1mole \[\equiv \frac{3}{2}\]mole ; 2 mole ? 3 mole \[V=3\times 22.4=67.2L\].You need to login to perform this action.
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