Question

From this combustion equation, 2CH22 + 3102 - 22H,0 + 2000, calculate the liters of

carbon dioxide produced when 16.9 grams of CH are combusted

Answer 1

Answer : The volume of
`CO_2` produced are, 26.7 liters.

Explanation :

First we have to calculate the moles of
`C_(10)H_(22)`

`\text{Moles of }C_(10)H_(22)=\frac{\text{Given mass }C_(10)H_(22)}{\text{Molar mass }C_(10)H_(22)}=(16.9g)/(142g/mol)=0.119mol`

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

The given combustion reaction is:

`2C_(10)H_(22)+31O_2\rightarrow 22H_2O+20CO_2`

From the balanced chemical reaction we conclude that,

As, 2 moles of
`C_(10)H_(22)` react to give 20 moles of
`CO_2`

So, 0.119 moles of
`C_(10)H_(22)` react to give
`(20)/(2)* 0.119=1.19` moles of
`CO_2`

Now we have to calculate the volume of
`CO_2` produced.

As we know that, 1 mole of gas occupies 22.4 L volume of gas.

As, 1 mole of
`CO_2` gas occupies 22.4 L volume of
`CO_2` gas.

So, 1.19 mole of
`CO_2` gas occupies 1.19 × 22.4 L = 26.7 L volume of
`CO_2` gas.

Therefore, the volume of
`CO_2` produced are, 26.7 liters.