A) an increase of 50 mL
B) an increase of 100 mL
C) an increase of 150 mL
D) a decrease of 50 mL
Correct Answer: A
Solution :
\[\underset{100\,mL}{\mathop{2P{{H}_{3}}(g)}}\,\xrightarrow{{}}2P(s)+\underset{150\,mL}{\mathop{3{{H}_{2}}(g)}}\,\] \[\therefore \] Volume change in the reaction = (150 -100) = 50 mL (increase)
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