If 20.5 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over?

The number of moles of excess reactant  left over will be, 0.089 moles.

Explanation : Given,

Mass of = 20.5 g

Mass of = 13.8 g

Molar mass of = 30 g/mol

Molar mass of = 32 g/mol

First we have to calculate the moles of and .

and,

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 2 mole of react with 1 mole of

So, 0.683 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of excess reactant  left over.

Number of moles of excess reactant  left over = Given moles - Required moles

Number of moles of excess reactant  left over = 0.431 mol - 0.342 mol

Number of moles of excess reactant  left over = 0.089 mol

Therefore, the number of moles of excess reactant  left over will be, 0.089 moles.

the answer is the first option

explanation:

just googled it m8. not that hard

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