A) 2 moles, 11 moles
B) 2 moles, 8 moles
C) 4 moles, 5 moles
D) 8 moles, 2 moles
Correct Answer: A
Solution :
[a] \[\underset{10}{\mathop{2S{{O}_{2}}}}\,+\underset{15}{\mathop{{{O}_{2}}}}\,\xrightarrow{{}}\underset{0}{\mathop{2S{{O}_{3}}}}\,\] |
\[10-2x~~~~~~~15-x~~~~~2x\] |
\[\therefore 2x=8\text{ }\,c=4\] |
Hence, remaining, \[S{{O}_{2}}=10-8=2\]moles, |
\[{{O}_{2}}=15-4=11\]moles |
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