Respuesta :
Answer : The mass of hydrogen gas consumed will be, 4.5 grams
Explanation : Given,
Mass of [tex]N_2[/tex] = 21 g
Mass of [tex]H_2[/tex] = 18 g
Molar mass of [tex]N_2[/tex] = 28 g/mole
Molar mass of [tex]H_2[/tex] = 2 g/mole
First we have to calculate the moles of [tex]N_2[/tex] and [tex]H_2[/tex].
[tex]\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{21g}{28g/mole}=0.75moles[/tex]
[tex]\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{18g}{2g/mole}=9moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]N_2(g)+3H_2(g)\rightarrow 2NH_3(g)[/tex]
From the given balanced reaction, we conclude that
As, 1 mole of [tex]N_2[/tex] react with 3 moles of [tex]H_2[/tex]
So, 0.75 moles of [tex]N_2[/tex] react with [tex]3\times 0.75=2.25[/tex] moles of [tex]H_2[/tex]
From this we conclude that, [tex]H_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]N_2[/tex] is a limiting reagent because it limits the formation of product.
The moles of hydrogen gas consumed = 2.25 mole
Now we have to calculate the mass of hydrogen gas consumed.
[tex]\text{Mass of }H_2=\text{Moles of }H_2\times \text{Molar mass of }H_2[/tex]
[tex]\text{Mass of }H_2=(2.25mole)\times (2g/mole)=4.5g[/tex]
Therefore, the mass of hydrogen gas consumed will be, 4.5 grams
