Question #a73d6

Let us look at the following equations

3N#O_2# + #H_2#O −→ 2HN#O_3# + NO.

As per the above equation 3 moles of N#O_2# are needed to form or get 2 moles of HN#O_3#.

in terms of molar mass , one mole of N#O_2# has mass 46.0 g and one mole of HN#O_3# has mass 63.0 g ( approx.)

3 mole of N#O_2# produces 2 moles of HN#O_3#

46 x 3 g of N#O_2# produces 2 x 63 g of HN#O_3#

138 g of N#O_2# produces 126 g of HN#O_3#

To get 1 g of HN#O_3# we will need ( 138g / 126g ) of N#O_2#

To get 1 g of HN#O_3# we will need 1.10g of N#O_2#

To get 63.0 g of HN#O_3# we will need 63 x 1.10 g of N#O_2#

69.3 g of N#O_2#

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)#).