Part A When nitrogen dioxide (NO2) from car exhaust water in the air, it forms n

Question

Part A When nitrogen dioxide (NO2) from car exhaust water in the air, it forms nitric acid (HNOs), which causes acid rain, and nitrogen oxide. with How of NO2 are needed to react with 0.250 mol of H207 ZAO, (9) + 11,0(1)--2HNO, (aq) + NO(s) molecules My Answers Give Up Part B How many grams of HNO, are produced when 60.0 g of NO2 completely reacts? Express your answer with the appropriate units. Part C How many grams of HNOs can be produced if 225 g of NO2 is mixed with 55.2 g of H,01? Express your answer with the appropriate units.

Explanation / Answer

PART A:

3 NO2(g)   +   H2O(l)     à     2HNO3(aq)   +   NO(g);

In the above reaction equation;

1 mole of H2O reacts with 3 moles of NO2

=> 0.250 mole of H2O reacts with (0.250 x 3) moles of NO2

=> 0.250 mole of H2O reacts with 0.750 moles of NO2

Now 1 mole of NO2 = 6.023 × 1023 molecules

0.750 mole of NO2 = 0.750 x 6.023 × 1023 molecules
                               = 4.52 × 1023 molecules

PART B:

3 NO2(g)   +   H2O(l)     à     2HNO3(aq)   +   NO(g);

In the above reaction equation;

3 moles of NO2 produces 2 moles of HNO3

=> 1 moles of NO2 produces 2/3 moles of HNO3

Molar mass of NO2 = 46 g/mol

So, 46 g of NO2 = 1 mol

1 g of NO2 = (1/46) mol

60 g of NO2 = (60/46) mol = 1.304 mol

Since, 1 moles of NO2 produces 2/3 moles of HNO3

=> 1.304 moles of NO2 produces 1.304 (2/3) moles of HNO3

=> 1.304 moles of NO2 produces 0.87 moles of HNO3

Molar mass of HNO3 = 63 g/mol

So, 1 mole of HNO3 = 63 g

0.87 moles of HNO3 = 0.87 x 63 g = 54.81 g

PART C:

3 NO2(g)   +   H2O(l)     à     2HNO3(aq)   +   NO(g);

In the above reaction equation;

3 moles of NO2 reacts with 1 mole of H2O to produce 2 moles of HNO3

Molar mass of NO2 = 46 g/mol

So, 46 g of NO2 = 1 mol

1 g of NO2 = (1/46) mol

225 g of NO2 = (225/46) mol = 4.89 mol                  (4.89/3 = 1.63)

Molar mass of H2O = 18 g/mol

So, 18 g of H2O = 1 mol

1 g of H2O = (1/18) mol

55.2 g of H2O = (55.2/18) mol = 3.07 mol

So, NO2 is the limiting reagent.

Now,

      3 moles of NO2 produces 2 moles of HNO3

=> 1 moles of NO2 produces 2/3 moles of HNO3

=> 4.89 moles of NO2 produces 4.89 (2/3) moles of HNO3

=> 4.89 moles of NO2 produces 3.26 moles of HNO3

Molar mass of HNO3 = 63 g/mol

So, 1 mole of HNO3 = 63 g

3.26 moles of HNO3 = 3.26 x 63 g = 205.38 g

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