I have another tedious stoicheometry . Let's see if we get the same answer?

Question

If .740 g of O3 reacts with .670g of NO, how many grams of NO2 will be produced? Which compound is the limiting reagent? Calculate the number of moles of the excess reagent remaining at the end of the reaction.



This one took me a while...Like almost 35 minutes. I couldnt get it together. It still has my brain in an bundle!

Answer 1

O3 = 16g x 3 = 48g/mol.
1 Mole O3 = 48g (÷ 4)
0∙02083333...Mole = 1g (x 0∙740)
0∙014516666.Mole = 0∙740g

NO = 14g + 16g = 30g/Mol.
1 Mole NO = 30g
0∙03333...Mole = 1g
0∙022333.. Mole = 0∙670g

If the O3 splits to give O2 + O,
then the reaction is 1:1, and O3 is the limiting reactant.

Excess NO:
= 0∙022333.. Mole of NO - 0∙014516666.Mole of O3
= 0∙0068666...Moles of NO.

If the O3 splits to give O2 + O, and splits again to give O + O + O, then the reaction is 3:1, and NO is the limiting reactant.

Excess O3:
= 2/3 x 0∙014516666.Mole
= 0∙009677777....Moles of O3.

Answer 2

1. Write the balance equation first.

O3 + NO ---> NO2 + O2

2. Determine the moles of each species.

mol O3 = (0.740g) / (48g/mol) = 0.0154 mol
mol NO = (0.670g) / (30g/mol) = 0.0223 mol

here, you'll know that the limiting reagent is O3.

3. Calculate for the weight of NO2 produced. Since your limiting reagent is O3, you'll know that only 0.154 mol will react based from the balanced equation. So,

weigth NO2 = 0.154 mol * 46g/mol = 7.084 grams

4. For the number of moles of the excess reagent, simply subtract mol O3 from mol NO. So,

mol of NO excess = 0.0223 - 0.0154 = 0.0069 mol