Question

Gaseous butane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 1.74 g of butane is mixed with 11. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answer 1

Answer : The mass of $CO_2$ produced will be, 5.3 grams.

Explanation : Given,

Mass of $C_4H_{10}$ = 1.74 g

Mass of $O_2$ = 11 g

Molar mass of $C_4H_{10}$ = 58 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $C_4H_{10}$ and $O_2$.

$\text{Moles of }C_4H_{10}=\frac{\text{Mass of }C_4H_{10}}{\text{Molar mass of }C_4H_{10}}=\frac{1.74g}{58g/mole}=0.03moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{11g}{32g/mole}=0.34moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_4H_{10}$ react with 13 mole of $O_2$

So, 0.03 moles of $C_4H_{10}$ react with $\frac{13}{2}\times 0.03=0.195$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C_4H_{10}$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$.

As, 2 moles of $C_4H_{10}$ react to give 8 moles of $CO_2$

So, 0.03 moles of $C_4H_{10}$ react to give $\frac{8}{2}\times 0.03=0.12$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$.

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.12mole)\times (44g/mole)=5.3g$

Therefore, the mass of $CO_2$ produced will be, 5.3 grams.