How many grams of NO2 are theoretically produced if we start with 1.20 moles of S and 9.90 moles of HNO3

How many grams of NO2 are theoretically produced if we start with 1.20 moles of S and 9.90 moles of HNO3?
Reaction: S + 6HNO3 → H2SO4 + 6NO2 + 2H2O

Don't just post the answer; explain how you got it.

Balanced equation: S + 6 HNO3 -> H2SO4 + 6 NO2 + 2 H2O

Since the reactants are already given us in moles, we can skip step 1 of limiting reactant problem, where we have to find the number of moles of each reactant.

2. Using the balanced equation, set up a mole ratio: Coefficients/Number of moles.

1 S/1.20 mol S = 6 HNO3/x

We pretend we don't know the moles of HNO3.

By crossing multiplying, we find x = 7.20 mol HNO3, which is less than 9.90 mol NHO3. This tells us we will have EXTRA HNO3, and that S is our limiting reactant.


3. Do a regular stoichiometry problem:

Molar ratio: 6 mol NO2 : 1 mol S

Molar mass NO2: 46.01 g/mol

1.20 mol S x 6 mol NO2/1 mol S x 46.01 g NO2/1 mol NO2 = 331.27 g NO2


Sig figs... 331 g NO2