Answer : The volume of [tex]CO_2[/tex] produced are, 26.7 liters.
Explanation :
First we have to calculate the moles of [tex]C_{10}H_{22}[/tex]
[tex]\text{Moles of }C_{10}H_{22}=\frac{\text{Given mass }C_{10}H_{22}}{\text{Molar mass }C_{10}H_{22}}=\frac{16.9g}{142g/mol}=0.119mol[/tex]
Now we have to calculate the moles of [tex]CO_2[/tex].
The given combustion reaction is:
[tex]2C_{10}H_{22}+31O_2\rightarrow 22H_2O+20CO_2[/tex]
From the balanced chemical reaction we conclude that,
As, 2 moles of [tex]C_{10}H_{22}[/tex] react to give 20 moles of [tex]CO_2[/tex]
So, 0.119 moles of [tex]C_{10}H_{22}[/tex] react to give [tex]\frac{20}{2}\times 0.119=1.19[/tex] moles of [tex]CO_2[/tex]
Now we have to calculate the volume of [tex]CO_2[/tex] produced.
As we know that, 1 mole of gas occupies 22.4 L volume of gas.
As, 1 mole of [tex]CO_2[/tex] gas occupies 22.4 L volume of [tex]CO_2[/tex] gas.
So, 1.19 mole of [tex]CO_2[/tex] gas occupies 1.19 × 22.4 L = 26.7 L volume of [tex]CO_2[/tex] gas.
Therefore, the volume of [tex]CO_2[/tex] produced are, 26.7 liters.