Question

Question 3 1pts If the percent yield for the following reaction is 75.0%, and 45.0 g of NO2 are consumed in the reaction, how many grams of nitric acid, HNO3(aq) are produced?_8 3 NO2(g) + H2003—2HNO3(aq) + NO(g)

Answer 1

Solution-

We can see that three moles of NO2 produce two moles of HNO3 using the equation

Now we find moles of NO2 were consumed.

The molar mass of NO2 = 46 g/mol

45/46 = 0.978 moles

Again the number of moles of HNO3 produced divide by 3 and multiply by 2

(0.978/3) x 2 = 0.652 moles of HNO3

but the yield is only 75% so let us divide it by 100 and multiply by 75

(0.652/100) x 75 = 0.489 moles of HNO3

The molar mass of HNO3 = 63 g/mol

0.489 x 63 = 30.8 g of HNO3 produced