Answer:
18.75 g of NH3.
Explanation:
The balanced equation for the reaction is given below:
4NH3 + 5O2 → 4NO + 6H2O
Next, we shall determine the masses of NH3 and O2 that reacted from the balanced equation.
This can be obtained as follow:
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Mass of NH3 from the balanced equation = 4 x 17 = 68 g
Molar mass of O2 = 16x2 = 32 g/mol
Mass of O2 from the balanced equation = 5 x 32 = 160 g
From the balanced equation above,
68 g if NH3 reacted with 160 g of O2.
Next, we shall determine the excess reactant. This can be obtained as follow:
From the balanced equation above,
68 g if NH3 reacted with 160 g of O2.
Therefore, 40 g of NH3 will react with = (40 × 160)/68 = 94.12 g of O2.
From the calculations made above, we can see that it will take a higher amount of O2 i.e 94.12g than what was given i.e 50g to react completely with 40 g of NH3.
Therefore, O2 is the limiting reactant and NH3 is the excess reactant.
Next we shall determine the mass of excess reactant that reacted. This can be obtained as follow:
From the balanced equation above,
68 g if NH3 reacted with 160 g of O2.
Therefore, Xg of NH3 will react with 50 g of O2 i.e
Xg of NH3 = (68 × 50)/160
Xg of NH3 = 21.25 g
Therefore, 21.25 g of NH3 (excess reactant) were consumed in the reaction.
Finally, we shall determine mass of the remaining excess reactant as follow:
Mass of excess reactant = 40 g
Mass of excess reactant that reacted = 21.25 g
Mass of excess reactant remainig =?
Mass of excess reactant remainig = (Mass of excess reactant) – (Mass of excess reactant that reacted)
Mass of excess reactant remainig
= 40 – 21.25
= 18.75 g
Therefore, the mass of excess reactant remaining is 18.75 g of NH3.
