Answer: The number of oxygen gas molecules initially present are [tex]3.409\times 10^{21}[/tex]
Explanation:
The balanced chemical reaction for the combustion of given hydrocarbon follows:
[tex]C_xH_y+(x+\frac{y}{4})O_2\rightarrow xCO_2+\frac{y}{2}H_2O[/tex]
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of carbon dioxide = 0.1497 g
Molar mass of carbon dioxide = 44 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of carbon dioxide}=\frac{0.1497g}{44g/mol}=3.403\times 10^{-3}mol[/tex]
Thus, [tex]x=3.402\times 10^{-3}mol[/tex]
Given mass of water = 0.0817 g
Molar mass of water = 18 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of water}=\frac{0.0817g}{18g/mol}=4.54\times 10^{-3}mol[/tex]
Thus, [tex]\frac{y}{2}=4.54\times 10^{-3}mol[/tex]
To calculate the number of moles of oxygen, we solve the equation for its coefficient by putting values of 'x' and 'y', we get:
[tex]\Rightarrow (3.402\times 10^{-3})+(\frac{4.54\times 10^{-3}}{2})=5.662\times 10^{-3}mol[/tex]
According to mole concept:
1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules.
So, [tex]5.662\times 10^{-3}[/tex] moles of oxygen will contain = [tex]5.662\times 10^{-3}\times 6.022\times 10^{23}=3.409\times 10^{21}[/tex] molecules.
Hence, the number of oxygen gas molecules initially present are [tex]3.409\times 10^{21}[/tex]
