Respuesta :
Answer: 1. Oxygen is the the limiting reactant.
2. 7.52%
Explanation:
The balanced chemical equation is:
[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(g)[/tex]
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} H_2=\frac{30.0g}{2g/mol}=15.0moles[/tex]
[tex]\text{Moles of} O_2=\frac{5.29g}{32g/mol}=0.165moles[/tex]
According to stoichiometry :
1 mole of [tex]O_2[/tex] require = 2 moles of [tex]H_2[/tex]
Thus 0.165 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.165=0.331moles[/tex] of [tex]NH_3[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
2. [tex]\text{Moles of} H_2=\frac{9.93g}{2g/mol}=4.96moles[/tex]
[tex]\text{Moles of} H_2O=\frac{6.72g}{18g/mol}=0.373moles[/tex]
As 2 moles of [tex]H_2[/tex] give = 2 moles of [tex]H_2O[/tex]
Thus 4.96 moles of [tex]H_2[/tex] give =[tex]\frac{2}{2}\times 4.96=4.96moles[/tex] of [tex]H_2O[/tex]
percentage yield = [tex]\frac{\text {actual yield}}{\text {theoretical yield}}=\frac{0.373}{4.96}\times 100=7.52\%[/tex]
Thus the percent yield for the reaction is 7.52%
