In one experiement 0.886 mole of NO is mixed with 0.503 mole of 0(2-). Calculate which two of the reactants?

Question

is the limiting reagent. Calculate also the number of moles of NO2 produced.

Let's See What you get....This time around. Chem Lovers

Answer 1

According to the equation:

2NO + O2 --> 2NO2

2 mol of NO reacts with 1 mol O2, so
0.886 mol NO reacts with 0.886/2 = 0.443 mol O2

We have 0.503 mol O2 > 0.443 mol. So O2 is in excess and NO is the limiting reactant. It reacts fully and produces 0.886 mol NO2 (because 2 mol NO give 2 mol NO2)

Answer 2

Assuming 2NO + O2 --> 2NO2
2 moles of NO is required for every 1 mole of O2
0.886/2 = 0.443 < 0.503, so the limiting reagent is NO, and 0.866 moles of NO2 will be produced.

Answer 3

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