f 24.7 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over? 2 NO (g) + O₂ (g) → 2 NO₂ (g)

0.02 moles of O₂ will be leftover.

Explanation:

The reaction is:

2NO(g) + O₂(g) → 2NO₂(g)    (1)

We have the mass of NO and O₂, so we need to find the number of moles:

From equation (1) we have that 2 moles of NO reacts with 1 mol of O₂ to produce 2 moles of NO₂, so the excess reactant is:

Hence, from above we can see that the excess reactant is O₂ since 0.41 moles react with 0.86 moles of NO and we have 0.43 moles in total for O₂.

The number of moles of excess reactant is:

Therefore, 0.02 moles of O₂ will be leftover.

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i think that the answer is c