Respuesta :
Answer:a) 11.34 g of ethane [tex](C_2H_6)[/tex] can be formed
b) [tex]C_2H_4[/tex] is the limiting reagent
c) 3.44 g of the excess reagent remains after the reaction is complete
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
1. [tex]\text{Moles of} H_2=\frac{4.21}{2}=2.10moles[/tex]
2. [tex]\text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles[/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] require 1 mole of [tex]H_2[/tex]
Thus 0.378 moles of [tex]C_2H_4[/tex] will require=[tex]\frac{1}{1}\times 0.378=0.378moles[/tex] of [tex]H_2[/tex]
Thus [tex]C_2H_4[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
moles of [tex]H_2[/tex] left = (2.10-0.378) = 1.72 moles
mass of [tex]H_2[/tex] left=[tex]moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g[/tex]
According to stoichiometry :
As 1 mole of [tex]C_2H_4[/tex] give = 1 mole of [tex]C_2H_6[/tex]
Thus 0.378 moles of [tex]C_2H_4[/tex] give =[tex]\frac{1}{1}\times 0.378=0.378moles[/tex] of [tex]C_2H_6[/tex]
Mass of [tex]C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g[/tex]
Thus 11.34 g of ethane is formed.
