Chemistry Honors Help? (stoichiometry)

Question

7) How many moles of nitric oxide, NO, are formed when 1.2 mol of nitrogen dioxide, NO2, reacts with excess water?

3NO2 (g) + H2O (l) --> 2HNO3 (aq) + NO (g)

82) If 60 grams of NO and 60 grams of O2 are available, what is the maximum amount of NO2 that can be produced?

Answer 1

7). Look at the correctly balanced equation:

3NO₂ (g) + H₂O (l) ==> 2HNO₃ (aq) + NO (g)

1.2 mol......xs........................................?..........

Using the mole ratio (stoichiometry) of 1 mol NO : 3 mol NO₂ (or it takes 3 mol NO₂ to make 1 mol NO)...

we can set up the following relationship (using dimensional analysis):

1.2 moles NO₂ x 1 mole NO / 3 moles NO₂ = 0.4 moles NO formed

82). Write a correctly balanced equation for the reaction taking place:

2NO(g) + O₂(g) ==> 2NO₂(g)

60 g........60 g.............?....... This is a problem about limiting reactants. Which reactant will run out first?

moles NO present = 60 g NO x 1 mol NO/30 g = 2 moles NO

moles O₂ present = 60 g O₂ x 1 mol O2/32 g = 1.875 moles O₂

NO is limiting and will run out first because it takes TWO moles of NO for each ONE mole of O₂ in the balanced equation.

Now we use the limiting reactant to calculate how much product (NO₂) can be formed:

2 moles NO x 2 moles NO₂ / 2 moles NO x 46 g/mole NO₂ = 92 g of NO₂ can be produced