The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to product. Consider the reaction: H2(g) + C2H4(g) → C2H6(g) If 8.150 g H2 is mixed with 9.330 g C2H4, calculate the theoretical yield (g) of C2H6 produced by the reaction.

Respuesta :

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