Respuesta :
Answer: a. 7.31 g
b. 20.4 %
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} H_2=\frac{1.29g}{2g/mol}=0.645moles[/tex]
[tex]\text{Moles of} N_2=\frac{9.55}{28g/mol}=0.341moles[/tex]
[tex]3H_2(g)+N_2(g)\rightarrow 2NH_3(g)[/tex]
According to stoichiometry :
3 moles of [tex]H_2[/tex] require = 1 mole of [tex]N_2[/tex]
Thus 0.645 moles of [tex]H_2[/tex] will require=[tex]\frac{1}{3}\times 0.645=0.215moles[/tex] of [tex]N_2[/tex]
Thus [tex]H_2[/tex] is the limiting reagent as it limits the formation of product and [tex]N_2[/tex] is the excess reagent.
As 3 moles of [tex]H_2[/tex] give = 2 moles of [tex]NH_3[/tex]
Thus 0.645 moles of [tex]H_2[/tex] give =[tex]\frac{2}{3}\times 0.645=0.430moles[/tex] of [tex]NH_3[/tex]
Mass of [tex]NH_3=moles\times {\text {Molar mass}}=0.430moles\times 17g/mol=7.31g[/tex]
Thus theoretical yield for this reaction under the given conditions is 7.31 g.
b) [tex]{\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%[/tex]
[tex]{\text {percentage yield}}=\frac{1.49g}{7.31g}\times 100\%=20.4\%[/tex]
The percent yield for this reaction under the given conditions is 20.4 %
