If 10g of NO react with 20g of O2, what is the maximum amount of NO2 that can be produced?

My answer is

Mol NO = 10/30 = 0.333 mols
Mol O2 = 20/32 = 0.625 mols

To react all the O2 you would need 2 NO*0.625 moles of NO/1mole O2 =1.25 O2? but, we just have 0.333 mols, so NO is your limiting reactant and O2 is in excess.

Are the units correct on the calculation; where I put 2 NO, should it be 2 moles NO * 0.625 moles NO/1mole O2?? Confused here.

What you have is mathematically correct; however, I think it is mixed up. You have made the factor (0.625/1 mol O2).

The factor is (2 mols NO/1 mol O2) and I would write, to convert O2 to NO (to use all of the O2) we would need the following NO.
mols NO needed = 0.625 mols O2 x (2 mols NO/1 mol O2) = 1.25 mols NO. However, we have only 0.333 mols NO, therefore, NO is the limiting reagent.

Here is what you are doing with this method. You ask yourself, how much NO do I need to react with 0.625 mols O2. Then you start with what you are given (that's 0.625 mols O2) and convert that using the dimensional method to mols NO.
Notice how the units cancel.
0.625 mols O2 x (2 mols NO/1 mol O2) = 0.625 x (2/1) = 1.25 mols NO.
This ALWAYS follows the pattern of
what is given x (factor) = what is wanted.
We are given O2, we multiply that by a factor and that changes it to what we want(mols NO).
Notice in the above that the mols O2 (bolded in the numerator and denominator cancel to leave a unitless number) and the italicized numerator (mols NO) is the unit we want to keep and the unit we want for the answer.

Changing feet to inches is the same kind of problem but we mix in the word chemistry and students think it's hard. Suppose we have 24 inches and we want to convert to feet. The factor is (1 foot = 12 inches)
We are given 24 inches so we follow the
given x factor = what we want
24 inches x (1 foot/12 inches) = 2 feet.
Note the inches (bolded) cancel and the expression leaves the unit we want to keep (feet) to stay so the answer has the unit of feet.

Your calculation for the number of moles of NO and O2 is correct. However, your understanding of the stoichiometry of the reaction seems to be a bit confused. Let me clarify it for you.

The balanced chemical equation for the reaction between NO and O2 to produce NO2 is:

2NO + O2 -> 2NO2

From the equation, we can see that 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2.

In your calculation, you correctly determined that you have 0.333 moles of NO and 0.625 moles of O2. Therefore, we need to consider the stoichiometry of the reaction to determine the limiting reactant.

To find the limiting reactant, we compare the moles of each reactant to their stoichiometric ratio in the balanced equation. In this case, the stoichiometric ratio is 2 moles of NO for every 1 mole of O2.

So, for every 1 mole of O2, we need 2 moles of NO. Therefore, since you have 0.333 moles of NO and only 0.625 moles of O2, the NO is the limiting reactant because you do not have enough NO to react with all the available O2.

Now, to find the maximum amount of NO2 that can be produced, we will use the limiting reactant, which is NO. Since the stoichiometry of the reaction tells us that 2 moles of NO are required to produce 2 moles of NO2, we can create an equivalence:

0.333 moles NO -> x moles NO2

Using this equivalence, we can see that the maximum amount of NO2 that can be produced is also 0.333 moles.

To convert moles to grams, you can use the molar mass of NO2, which is 46 g/mol. Therefore, the maximum amount of NO2 that can be produced is:

Mass of NO2 = 0.333 moles x 46 g/mol = 15.348 g

So, the maximum amount of NO2 that can be produced is approximately 15.348 grams.

I hope this explanation helps clarify your confusion. Let me know if you have any more questions!