How many grams of HNO3 can be prepared from the reaction of 92.0g of NO2 with 36.0g of H2O?

Question

Reaction: 3NO2+H2O--->2HNO3+NO

Answer 1

Well since you already have the equation what you need to to is work out the number of moles of each NO2 and H2O to find your limiting reagent (the substance that will run out first).

n(NO2)=92.0/(14.01+2x16)=2 moles

n(H2O)=36/(2x1.008+16)=2 moles

However as you need 3 moles of NO2 for every mole of H2O the NO2 is your limiting reagent.

The next step is to find the number of moles of HNO3 produced. For every  three moles of NO2, two moles of HNO3 are produced.

This means that in this equation n(HNO3)=2/3 x 2 =1 1/3 moles

In calculating the mass you times the number of moles by the molar mass of HNO3. Thus:

m(HNO3)= 1 1/3 x (1.008+14.01+3x16)=84.0g (to three significant figures as in the question)  

Answer 2

3 ( 46 grams/mole ) = 138 grams of NO2 react with 1 mole ( 18 grams/mole ) = 18 grams of water to produce  2 moles ( 63grams/mole ) = 126 grams of HNO3

92 grams of NO2 will require

138/18 = 92/X = 12 grams  so H2O is in excess  no worries....138 grams of NO2 gives 126 grams of HNO3   so

138/126 = 92/X = 84 grams = 84grams = 84 /63 = 1.33 moles

92 grams of NO2 = 2 moles

3moles/2 moles = 2 moles /1.33 moles