Answer: The maximum mass of
produced is, 61.0 grams.
Explanation : Given,
Mass of
= 90.6 g
Mass of
= 90.6 g
Molar mass of
= 17 g/mol
Molar mass of
= 32 g/mol
First we have to calculate the moles of
and
.
![\text{Moles of }NH_3=\frac{\text{Given mass }NH_3}{\text{Molar mass }NH_3}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DNH_3%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DNH_3%7D%7B%5Ctext%7BMolar%20mass%20%7DNH_3%7D)
![\text{Moles of }NH_3=\frac{90.6g}{17g/mol}=5.33mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DNH_3%3D%5Cfrac%7B90.6g%7D%7B17g%2Fmol%7D%3D5.33mol)
and,
![\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DO_2%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%20%7DO_2%7D%7B%5Ctext%7BMolar%20mass%20%7DO_2%7D)
![\text{Moles of }O_2=\frac{90.6g}{32g/mol}=2.83mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DO_2%3D%5Cfrac%7B90.6g%7D%7B32g%2Fmol%7D%3D2.83mol)
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
![4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)](https://tex.z-dn.net/?f=4NH_3%28g%29%2B5O_2%28g%29%5Crightarrow%204NO%28g%29%2B6H_2O%28g%29)
From the balanced reaction we conclude that
As, 5 mole of
react with 4 mole of ![NH_3](https://tex.z-dn.net/?f=NH_3)
So, 2.83 moles of
react with
moles of ![NH_3](https://tex.z-dn.net/?f=NH_3)
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
From the reaction, we conclude that
As, 5 mole of
react to give 6 mole of ![H_2O](https://tex.z-dn.net/?f=H_2O)
So, 2.83 mole of
react to give
mole of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Now we have to calculate the mass of ![H_2O](https://tex.z-dn.net/?f=H_2O)
![\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DH_2O%3D%5Ctext%7B%20Moles%20of%20%7DH_2O%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DH_2O)
Molar mass of
= 18 g/mole
![\text{ Mass of }H_2O=(3.39moles)\times (18g/mole)=61.0g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DH_2O%3D%283.39moles%29%5Ctimes%20%2818g%2Fmole%29%3D61.0g)
Therefore, the maximum mass of
produced is, 61.0 grams.