Question

could someone please check this for me?

If 3.5 moles of nitrogen monoxide (NO) react with 6.0 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

2NO + O2 -> 2NO2

3.5 mol NO x [2 mol NO2]/[2 mol NO] = 3.5 mol NO2

6.0 mol O2 x [2 mol NO2]/[1 mol O2] = 12 mol NO2

3.5 mol NO2 x [1 mol O2]/[2 mol NO2] = 1.75 mol O2

6.0 mol O2 - 1.75 mol O2 = 4.25 mol O2

3.5 moles of the product can be formed. 4.25 moles of the excess reactant will be left over.

Answer 2

it seemed too easy while i was working on it, so i'm concerned that i did it all wrong lol

Answer 3

It all look perfect to me as your limited reactant is nitrogen monoxide.

Answer 4

so i'm right? yey :3 thanks

Answer 5

You are right. Chemistry don't have to be hard when you're good at it ;)

Answer 6

Don't know if you do the same but I always like to set up this equation:
For a reaction a A + b B -> c C + d D there most folliow:
\[\large \frac{ n(A) }{ a }=\frac{ n(B) }{ b }=\frac{ n(C) }{ c }=\frac{ n(D) }{ d }\]
So for your question:
\[\large \frac{ n(OH) }{ 2 }=\frac{ n(O _{2}) }{ 1 }=\frac{ n(NO _{2}) }{ 2 }\]
don't know if that can help doing it even more easy than it already is? :)