Question

$C_2{H}_{5}OH+3O_2\to 2C{O}_2+3H_{2}O$
According to the reaction above, if $5$ moles of $C{O}_2$ are produced, then which of the following is true? $(MW$ of $C_2{H}_{5}OH=46)$

• A. $115g$ of $C_2{H}_{5}OH$ are consumed
• B. $58g$ of $C_2{H}_{5}OH$ are consumed
• C. $192g$ of $O_2$ are consumed
• D. $96g$ of $O_2$ are consumed

Answer 1

The correct option is A $115g$ of $C_2{H}_{5}OH$ are consumed

$C_2{H}_{5}OH$ $+$ $3O_2$ $=$ $2C{O}_2$ $+$ $3H_{2}O$

From the above equation it is seen that two moles of Carbon Dioxide are produced from one mole of Ethanol. Molecular weight of Carbon Dioxide is $44$ grams.

Thus, two moles of $C{O}_2$ weighs ($2$$\ast$$44$) grams, that is $88grams$.

Five moles of $C{O}_2$ weighs ($5$$\ast$$44$) grams, that is $220grams$.

One mole of Ethanol weighs $46grams$.

Thus, from the equation above-

$88grams$ of $C{O}_2$ is obtained from $46grams$ of $C_2{H}_{5}OH$

$1gram$ of $C{O}_2$ is obtained from $\frac{46}{88}$$grams$ of $C_2{H}_{5}OH$

$220grams$ of $C{O}_2$ is obtained from $\frac{46}{88}$$\ast$$220$ $grams$, that is $115grams$ of $C_2{H}_{5}OH$

Option A is the correct answer.