Question #a73d6

2 Answers
Dec 11, 2014

Let us look at the following equations

3NO_2 + H_2O −→ 2HNO_3 + NO.

As per the above equation 3 moles of NO_2 are needed to form or get 2 moles of HNO_3.

in terms of molar mass , one mole of NO_2 has mass 46.0 g and one mole of HNO_3 has mass 63.0 g ( approx.)

3 mole of NO_2 produces 2 moles of HNO_3

46 x 3 g of NO_2 produces 2 x 63 g of HNO_3

138 g of NO_2 produces 126 g of HNO_3

To get 1 g of HNO_3 we will need ( 138g / 126g ) of NO_2

To get 1 g of HNO_3 we will need 1.10g of NO_2

To get 63.0 g of HNO_3 we will need 63 x 1.10 g of NO_2

69.3 g of NO_2

Dec 11, 2014

The answer is 83g.

Starting from the balanced chemical equation

3NO_2 + H_2O -> 2HNO_3 + NO

one can see that we have a 3:2 mole ratio between NO_2 and H_2O; that is, for every 3 moles of NO_2 that react, 2 moles of HNO_3 are formed.

Now, starting from the given mass of nitric acid, m_(HNO_3) = 75g, one can determine the number of moles formed in the reaction and, by working backwards, determine how many moles of NO_2 were used. So,

n_(HNO_3) = (mass_(HNO_3))/(molarmass) = (75g)/(63 g/(mol)) = 1.2 moles (knowing that nitric acid's molar mass is 63g/(mol)).

Therefore, the number of NO_2 moles used is

n_(NO_2) = 3/2 * 1.2 = 1.8 moles

Using the same conversion between moles and mass, one can determine that

m_(NO_2) = 1.8 mol es * 46 g/(mol) = 83g (NO_2's molar mass is 46 g/(mol)).