Question

Answer the following questions by referring to the equation below:
2 C6H10 + 17 O2  12 CO2 + 10 H2O

If 35 grams of C6H10 and 45 grams of oxygen are reacted,
i) which one is the limiting reagent?
ii) how many grams of carbon dioxide will be formed?
iii) how much of the excess reagent is left over after the reaction is complete?
iv) if 35 grams of carbon dioxide are formed, calculate the percent yield of this reaction.

Answer 1

$1mol C_6H_{10}$ reacts with $8.5mol O_2$

Moles of $C_6H_{10}=\frac{35}{82}=0.426mol$

Moles of $O_2=\frac{45}{32}=1.406mol$

As $0.165mol C_6H_{10}$ is required to completely react with $1.406mol O_2$

So $O_2$ is a limiting reagent

Moles of $CO_2$ formed when $1.406mol O_2$ reacted$=1.406\times\frac{12}{17}=0.992mol$

Hence mass of $CO_2$ formed$=0.992\times44=43.648g$

Moles of excess $C_6H_{10}$ left over after reaction$=0.426-0.165=0.261mol$

If $35g CO_2$ is formed then percent yield$\frac{35}{43.648}\times100=80.18$