1. Nitric acid is produced via the following equation: 3 NO2 + H2O = 2HNO3 + NO. How many grams of NO2 are required to produce 7.5 grams of HNO3?
To solve this problem, we need to use stoichiometry – the calculation of the quantities of reactants and products in chemical reactions.
First, let's determine the molar mass of the compounds involved:
- NO2 (Nitrogen Dioxide) has a molar mass of 46 grams/mol
- HNO3 (Nitric Acid) has a molar mass of 63 grams/mol
Now, let's convert the given mass of HNO3 to moles. We can use the formula:
moles = mass (g) / molar mass (g/mol)
moles of HNO3 = 7.5 g / 63 g/mol
moles of HNO3 ≈ 0.119 mol
According to the balanced equation, the stoichiometric ratio between NO2 and HNO3 is 3:2. This means that for every 3 moles of NO2, we get 2 moles of HNO3.
Now, we can set up a proportion using the stoichiometric ratio:
(0.119 mol HNO3 / 2 mol HNO3) = (x mol NO2 / 3 mol NO2)
Solving for x, we find:
x = (0.119 mol HNO3 * 3 mol NO2) / 2 mol HNO3
x ≈ 0.179 mol NO2
Finally, let's convert moles of NO2 to grams using the molar mass:
mass (g) = moles * molar mass
mass (g) = 0.179 mol * 46 g/mol
mass (g) ≈ 8.234 g
Therefore, to produce 7.5 grams of HNO3, approximately 8.234 grams of NO2 is required.
Dear Brett,
molar mass of the HNO3= 1+14+16x3
=63
number of moles of the HNO3 produced in the reaction = 7.5/63
= 0.119
According to the equation provided 3 moles of NO2 will produce 2 moles of HNO3
Thus,number of moles of NO2 required=
0.119/2x3
= 0.179
mass of N02 required=
0.179x(14+32)
= 8.214g
Therefore, 8.214 grams of N02 is required
Hope it helps:)