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 [tex]H_2[/tex]
[tex]\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles[/tex]
b) moles of [tex]C_2H_4[/tex]
[tex]\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles[/tex]
[tex]H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)[/tex]
According to stoichiometry :
1 mole of [tex]C_2H_4[/tex] combine with 1 mole of [tex]H_2[/tex]
Thus 0.33 mole of [tex]C_2H_4[/tex] will combine with =[tex]\frac{1}{1}\times 0.33=0.33[/tex] mole of [tex]H_2[/tex]
Thus [tex]C_2H_4[/tex] is the limiting reagent as it limits the formation of product.
As 1 mole of [tex]C_2H_4[/tex] give = 1 mole of [tex]C_2H_6[/tex]
Thus 0.33 moles of [tex]C_2H_4[/tex] give =[tex]\frac{1}{1}\times 0.33=0.33moles[/tex] of [tex]C_2H_6[/tex]
Mass of [tex]C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g[/tex]
Thus theoretical yield (g) of [tex]C_2H_6[/tex] produced by the reaction is 9.9 grams