Question

HELP ME <3

If 5.8 moles of nitrogen monoxide (NO) react with 10.0 moles of oxygen gas (O2), how many moles of the product can be formed and how many moles of the excess reactant will be left over when the reaction is complete? Show all of your work.

unbalanced equation: NO + O2 --> NO2

Answer 1

The balanced equation is :

. . . . . . .2 NO . . . + . . . O2 . . . ---> . . . 2 NO2

. . . . . 2 moles . . . . . . . . . . . . . . . . . . 2 moles



NO : k1 = n(NO) / 2 = 2,9

O2 : k2 = n(O2) / 2 = 5

k1 < k2 . . . then NO is the reagent limiting


According the equation :

. . . . . . 2 mole of NO gives 2 moles of NO2
. .then 5,8 moles of NO gives 5,8 moles of NO2


According the equation :

. . . . . . 2 mole of NO requires 1 mole of O2
. .then 5,8 moles of NO requires 5,8 / 2 = 2,9 moles of O2

There is 10 - 2,9 = 7,1 moles of O2

Answer 2

The balanced chemical equation would be as follows:

2NO + O2 --> 2NO2

We are given the amount of reactants to be used for the reaction. We use these amounts to determine first which is the limiting reactant. The ratio of the reactants from the reaction is 2:1. Therefore, the limiting reactant would be nitrogen monoxide since it can be consumed completely. Calculations are as follows:

5.8 mol NO ( 2 mol NO2 / 2 mol NO ) = 5.8 mol NO2 is produced
10.0 mol O2 - (5.8 mol NO) ( 1 mol O2 / 2 mol NO ) = 7.1 mol O2 excess.