If 16.00 g of O₂ reacts with 80.00 g NO, how many grams of NO₂ are produced? (enter only the value, round to whole number)

46 g

Explanation:

The balanced equation of the reaction between O and NO is

2 NO  +  O₂  ⇔  2 NO₂

Now, you need to find the limiting reagent.  Find the moles of each reactant and divide the moles by the coefficient in the equation.

NO:  (80 g)/(30.006 g/mol) = 2.666 mol

        (2.666 mol)/2 = 1.333

O₂:  (16 g)/(31.998 g/mol) = 0.500 mol

     (0.500 mol)/1 = 0.500 mol

Since O₂ is smaller, this is the limiting reagent.

The amount of NO₂ produced will depend on the limiting reagent.  You need to look at the equation to determine the ratio.  For every mole of O₂ reacted, 2 moles of NO₂ are produced.

To find grams of NO₂ produced, multiply moles of O₂ by the ratio of NO₂ to O₂.  Then, convert moles of NO₂ to find grams.

0.500 mol O₂ × (2 mol NO₂/1 mol O₂) = 1.000 mol NO₂

1.000 mol × 46.005 g/mol = 46.005 g

You will produce 46 g of NO₂.

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