Answer: 44.5 g of [tex]H_2O[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} NH_3=\frac{65.8}{17}=3.87moles[/tex]
[tex]\text{Moles of} O_2=\frac{65.8}{32}=2.06moles[/tex]
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]
According to stoichiometry :
5 moles of [tex]O_2[/tex] require 4 moles of [tex]NH_3[/tex]
Thus 2.06 moles of [tex]O_2[/tex] will require=[tex]\frac{4}{5}\times 2.06=1.65moles[/tex] of [tex]NH_3[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]NH_3[/tex] is the excess reagent.
As 5 moles of [tex]O_2[/tex] give = 6 moles of [tex]H_2O[/tex]
Thus 2.06 moles of [tex]O_2[/tex] give =[tex]\frac{6}{5}\times 2.06=2.47moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]H_2O=moles\times {\text {Molar mass}}=2.47moles\times 18g/mol=44.5g[/tex]
Thus 44.5 g of [tex]H_2O[/tex] will be produced from the given masses of both reactants.
