Answer:
A. 19.2 g of O2.
B. 3.79 g of N2.
C. 54 g of H2O.
Explanation:
The balanced equation for the reaction is given below:
4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)
Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.
This is illustrated below:
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 = 3 x 32 = 96 g
Molar mass of N2 = 2x14 = 28 g/mol
Mass of N2 from the balanced equation = 2 x 28 = 56 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 6 x 18 = 108 g
Summary:
From the balanced equation above,
68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.
A. Determination of the mass of O2 needed to react with 13.6 g of NH3.
This is illustrated below:
From the balanced equation above,
68 g of NH3 reacted with 96 g of O2.
Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.
Therefore, 19.2 g of O2 are needed for the reaction.
B. Determination of the mass of N2 produced when 6.50 g of O2 react.
This is illustrated below:
From the balanced equation above,
96 g of O2 reacted to produce 56 g of N2.
Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.
Therefore, 3.79 g of N2 were produced from the reaction.
C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.
This is illustrated below:
From the balanced equation above,
68 g of NH3 reacted to 108 g of H2O.
Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.
Therefore, 54 g of H2O were obtained from the reaction.
