Answer: 0.0 grams
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
a) moles of butane
[tex]\text{Number of moles}=\frac{5.2g}{58.12g/mol}=0.09moles[/tex]
b) moles of oxygen
[tex]\text{Number of moles}=\frac{32.6g}{32g/mol}=1.02moles[/tex]
[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]
According to stoichiometry :
2 moles of butane require 13 moles of [tex]O_2[/tex]
Thus 0.09 moles of butane will require =[tex]\frac{13}{2}\times 0.09=0.585moles[/tex] of [tex]O_2[/tex]
Butane is the limiting reagent as it limits the formation of product and oxygen is present in excess as (1.02-0.585)=0.435 moles will be left.
Thus all the butane will be consumed and 0.0 grams of butane will be left.
