Answer: 61.3 g of [tex]H_2O[/tex] will be produced from the given masses of both reactants.
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]NH_3[/tex]
[tex]\text{Number of moles}=\frac{91.0g}{17g/mol}=5.35moles[/tex]
b) moles of [tex]O_2[/tex]
[tex]\text{Number of moles}=\frac{91.0g}{32g/mol}=2.84moles[/tex]
[tex]4NH_3+5O_2\rightarrow 4NO+6H_2O[/tex]
According to stoichiometry :
5 moles of [tex]O_2[/tex] require 4 moles of [tex]NH_3[/tex]
Thus 2.84 moles of [tex]O_2[/tex] require=[tex]\frac{4}{5}\times 2.84=2.27moles[/tex] of [tex]NH_3[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product.
As 5 moles of [tex]O_2[/tex] give = 6 moles of [tex]H_2O[/tex]
Thus 2.84 moles of [tex]O_2[/tex] give =[tex]\frac{6}{5}\times 2.84=3.41moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text {Molar mass}}=3.41moles\times 18g/mol=61.3 g[/tex]
Thus 61.3 g of [tex]H_2O[/tex] will be produced from the given masses of both reactants.
