Answer: 9.9 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 ethane
[tex]\text{Number of moles}=\frac{19.2g}{30g/mol}=0.64moles[/tex]
b) moles of oxygen
[tex]\text{Number of moles}=\frac{35g}{32g/mol}=1.09moles[/tex]
[tex]2C_2H_6+7O_2\rightarrow 4CO_2+6H_2O[/tex]
According to stoichiometry :
7 moles of oxygen require 2 moles of ethane
Thus 1.09 moles of oxygen require =[tex]\frac{2}{7}\times 1.09=0.31moles[/tex] of ethane
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product. Ethane is the excess reagent.
Ethane left = (0.64-0.31) = 0.33 moles
Mass of [tex]ethane=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9[/tex]
Thus 9.9 grams of ethane that could be left over by the chemical reaction
