Answer: 43.5g of carbon dioxide it produce at 95% efficiency
Explanation:
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
To calculate the moles, we use the equation:
moles of butane =[tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{15.0g}{58g/mol}=0.26moles[/tex]
According to stoichiometry:
2 moles of butane gives = 8 moles of carbon dioxide
Thus 0.26 moles of butane gives= [tex]\frac{8}{2}\times 0.26=1.04[/tex] moles of carbon dioxide
Mass of [tex]CO_2[/tex] produced=[tex]moles\times {\text {Molar mass}}=1.04\times 44=45.8g[/tex]
As yield of the reaction is 95%, the amount of [tex]CO_2[/tex] produced =[tex]\frac{95}{100}\times 45.8=43.5g[/tex]
Thus 43.5g of carbon dioxide it produce at 95% efficiency
