Respuesta :
Answer:
60.9 g
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Given: For [tex]NH_3[/tex]
Given mass = 90.1 g
Molar mass of [tex]NH_3[/tex] = 17.031 g/mol
Moles of [tex]NH_3[/tex] = 90.1 g / 17.031 g/mol = 5.2904 moles
Given: For [tex]O_2[/tex]
Given mass = 90.1 g
Molar mass of [tex]O_2[/tex] = 31.9988 g/mol
Moles of [tex]O_2[/tex] = 90.1 g / 31.9988 g/mol = 2.8157 moles
According to the given reaction:
[tex]4NH_3_{(g)}+5O_2_{(g)}\rightarrow 4NO_{(g)}+6H_2O_{(g)}[/tex]
4 moles of [tex]NH_3[/tex] react with 5 moles of [tex]O_2[/tex]
1 mole of of [tex]NH_3[/tex] react with 5/4 moles of [tex]O_2[/tex]
5.2904 moles of [tex]NH_3[/tex] react with [tex]\frac {5}{4}\times 5.2904[/tex] mole of [tex]O_2[/tex]
Moles of [tex]O_2[/tex] = 6.613 moles
Available moles of [tex]O_2[/tex] = 2.8157 moles
Limiting reagent is the one which is present in small amount. Thus, [tex]O_2[/tex] is limiting reagent. (2.8157 < 6.613)
The formation of the product is governed by the limiting reagent. So,
5 moles of [tex]O_2[/tex] gives 6 moles of [tex]H_2O[/tex]
1 mole of [tex]O_2[/tex] gives 6/5 moles of [tex]H_2O[/tex]
2.8157 moles of [tex]O_2[/tex] gives [tex]\frac {6}{5}\times 2.8157[/tex] moles of [tex]H_2O[/tex]
Moles of water = 3.37884 moles
Molar mass of [tex]H_2O[/tex] = 18.0153 g/mol
Mass of [tex]H_2O[/tex] = Moles × Molar mass = 3.37884 × 18.0153 g = 60.9 g
