JEE MAIN - Chemistry (2019 - 9th January Evening Slot - No. 22)
For the following reaction, in the mass of water produced from 445 g of C57H110O6 is :
2C57H110O6(s) + 163 O2(g) $$ \to $$ 114 CO2(g) + 110 H2O(l)
2C57H110O6(s) + 163 O2(g) $$ \to $$ 114 CO2(g) + 110 H2O(l)
490 g
445 g
495 g
890 g
Explanation
moles of C57H110O6(s) = $${{445} \over {890}}$$ = 0.5 moles
2C57H110O6(s) + 163 O2(g) $$ \to $$ 114 CO2(g) + 110 H2O(l)
nH2O = $${{110} \over 4}$$ = $${{55} \over 2}$$
mH2O = $${{55} \over 2}$$ $$ \times $$ 18
= 495 gm
2C57H110O6(s) + 163 O2(g) $$ \to $$ 114 CO2(g) + 110 H2O(l)
nH2O = $${{110} \over 4}$$ = $${{55} \over 2}$$
mH2O = $${{55} \over 2}$$ $$ \times $$ 18
= 495 gm
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