A) 490 g
B) 445 g
C) 495 g
D) 890 g
Correct Answer: C
Solution :
\[2{{C}_{2}}{{H}_{110}}{{O}_{6}}(s)+{{1630}_{2}}(g)\to 114C{{O}_{2}}(g)+110{{H}_{2}}P(l)\] \[\frac{Moles\,of\,{{C}_{57}}{{H}_{110}}{{O}_{6}}}{2}=\frac{Moles\,of\,{{H}_{2}}O}{110}\] \[\frac{\frac{445}{890}}{2}=\frac{\frac{Mass\,of\,{{H}_{2}}O}{18}}{110}\] \[Mass\,of\,{{H}_{2}}O=495g.\]You need to login to perform this action.
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