Answer:
5.74g of NO
Step-by-step explanation:
Step 1:
The balanced equation for the reaction. This is given below:
4NH3 + 5O2 —> 4NO + 6H2O
Step 2:
Determination of the masses of NH3 and O2 that reacted and the mass of NO produced from the balanced equation. This is illustrated below:
Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol
Mass of NH3 from the balanced equation = 4 x 17 = 68g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 5 x 32 = 160g
Molar Mass of NO = 14 + 16 = 30g/mol
Mass of NO from the balanced equation = 4 x 30 = 120g
From the balanced equation above,
68g of NH3 reacted with 160g of O2 to produce 120g of NO.
Step 3:
Determination of the limiting reactant.
We need to determine the limiting because it will be used to calculate the maximum yield of the reaction. This is illustrated below:
From the balanced equation above,
68g of NH3 reacted with 160g of O2.
Therefore, 3.25g of NH3 will react with = (3.25 x 160)/68 = 7.65g of O2.
From the simple illustration above, we can see that lesser mass of O2 is needed to react with 3.25g of NH3. Therefore, NH3 is the limiting reactant while O2 is the excess reactant.
Step 4:
Determination of the mass of NO produced from the reaction.
In this case the limiting reactant will be used because all of it were used in the reaction.
The limiting reactant is NH3.
From the balanced equation above,
68g of NH3 reacted to produce 120g of NO.
Therefore, 3.25g of NH3 will react to produce = (3.25 x 120)/68 = 5.74g of NO.
From the calculations made above, 5.74g of NO is produced.