Question

If $88g$ of $C_3{H}_8$ and $160g$ of $O_2$ are allowed to react to form $C{O}_2$ and $H_{2}O$, how many grams of $C{O}_2$ will be formed?

• A. $33g$
• B. $66g$
• C. $132g$
• D. $264g$
• E. None of the above

Answer 1

Answer: (C)

$132g$

$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$

The molar masses of $C_3{H}_8,O_2,C{O}_2$ are $44g/mol,$ $32g/mol$ and $44g/mol$ respectively.

The number of moles is the ratio of mass to molar mass.

The number of moles of $C_3{H}_8=\frac{88g}{44g/mol}=2mol$

Number of moles of $O_2=\frac{160g}{32g/mol}=5mol$

2 moles of propane will react with $2\times 5=10$ moles of oxygen, however, only 5 moles of oxygen are present.

Hence, oxygen is the limiting reagent and propane is the excess reagent.

5 moles of oxygen will give 3 moles of $C{O}_2$.

Mass of $C{O}_2=3mol\times 44g/mol=132g$