Answer:
A. 3.65 g of H2O.
B. 26.33 g of NO.
C. 7.53 g of HNO3.
Explanation:
The balanced equation for the reaction is given below:
3NO2(g) + H2O(l) —> 2HNO3(aq) + NO(g)
Next, we shall determine the masses of NO2 and H2O that reacted and the masses of HNO3 and NO produced from the balanced equation.
This is illustrated below:
Molar mass of NO2 = 14 + (16x2) = 46 g/mol
Mass of NO2 from the balanced equation = 3 x 46 = 138 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol
Mass of H2O from the balanced equation = 1 x 18 = 18 g
Molar mass of HNO3 = 1 + 14 + (16x3) = 63 g/mol
Mass of HNO3 from the balanced equation = 2 x 63 = 126 g.
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO from the balanced equation = 1 x 30 = 30 g.
Summary:
From the balanced equation above,
138g of NO2 reacted with 18 g of H2O to produce 126 g of HNO3 and 30 g of NO.
A. Determination of the mass of H2O required to react with 28.0 g
of NO2.
This is illustrated below:
From the balanced equation above,
138g of NO2 reacted with 18 g of H2O.
Therefore, 28 g if NO2 will react with = (28 x 18)/138 = 3.65 g of H2O.
Therefore, 3.65 g of H2O is needed for the reaction.
B. Determination of the mass of NO produced from 15.8 g of H2O.
This is illustrated below:
From the balanced equation above,
18 g of H2O reacted to produce 30 g of NO.
Therefore, 15.8 g of H2O will react to produce = (15.8 x 30)/18 = 26.33 g of NO.
Therefore, 26.33 g of NO were produced from the reaction.
C. Determination of the mass of HNO3 produced from 8.25 g of NO2.
This is illustrated below:
From the balanced equation above,
138g of NO2 reacted to produce 126 g of HNO3.
Therefore, 8.25 g of NO2 will react to produce = (8.25 x 126)/138 = 7.53 g of HNO3.
Therefore, 7.53 g of HNO3 were produced from the reaction.
