how many grams of NO2(subscript), if 0.740 g of O3(subscript) reacts with 0.670 g of NO in this reaction:

O3(subscript) + NO = O2(subscript) + NO2(subscript)
please let me see the solution please... thanks..
identify the limiting reactant

You need to calculate the number of moles of each reactant you have. The reactant with the least moles will be the limiting reactant. Then you can use this figure to work out the number of moles of product you will get, and use this to work out the mass of product you will get.

Let me see your working and I can tell if I agree with your answer.

I'm not sure of my answer which is 0.69 grams of NO2 produced.
Then my limiting reactant is O3, which has only 0.015 moles.
Where the NO has 0.022 moles...
Is it correct? How do I check if I got the right answer?

Yes. I think it's correct, however I got 0.71g of NO2.

Maybe you have rounded up in several places during your answer, this could account for the discrepancy.

However, I think your method is right, you just need to be careful not to round off until you get the final answer.

Yes, I rounded it off.. so my method is right, I just don't have to round it off. Well, thanks a lot for your time!
I appreciate it. Thanks! :)