Respuesta :
The balanced chemical reaction is:
4NH3(g) + 5O2(g) = 4NO(g) + 6H2O
We use the reaction and the amount of the reactant , NH3, to determine the amount of oxygen needed. We do as follows:
200 g NH3 ( 1 mol / 17.04 g) (5 mol O2 / 4 mol NH3) (32 g / 1 mol) = 469.48 g O2 needed
4NH3(g) + 5O2(g) = 4NO(g) + 6H2O
We use the reaction and the amount of the reactant , NH3, to determine the amount of oxygen needed. We do as follows:
200 g NH3 ( 1 mol / 17.04 g) (5 mol O2 / 4 mol NH3) (32 g / 1 mol) = 469.48 g O2 needed
Answer : The mass of oxygen needed are 470.592 grams.
Solution : Given,
Mass of [tex]NH_3[/tex] = 200.0 g
Molar mass of [tex]NH_3[/tex] = 17 g/mole
Molar mass of [tex]O_2[/tex] = 32 g/mole
First we have to calculate the moles of [tex]NH_3[/tex].
[tex]\text{ Moles of }NH_3=\frac{\text{ Mass of }NH_3}{\text{ Molar mass of }NH_3}=\frac{200.0g}{17g/mole}=11.765moles[/tex]
Now we have to calculate the moles of [tex]O_2[/tex]
The balanced chemical reaction is,
[tex]4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)[/tex]
From the reaction, we conclude that
As, 4 mole of [tex]NH_3[/tex] react with 5 mole of [tex]O_2[/tex]
So, 11.765 moles of [tex]NH_3[/tex] react with [tex]\frac{11.765}{4}\times 5=14.706[/tex] moles of [tex]O_2[/tex]
Now we have to calculate the mass of [tex]O_2[/tex]
[tex]\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2[/tex]
[tex]\text{ Mass of }O_2=(14.706moles)\times (32g/mole)=470.592g[/tex]
Therefore, the mass of oxygen needed are 470.592 grams.
