If 50.0 g of nitrogen dioxide and 15.0 g water react according to the following equation, how many grams of nitrogen monoxide will form?

10.87 g of NO.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3NO2 + H2O —> 2HNO3 + NO

Next, we shall determine the masses of NO2 and H2O that reacted and the mass NO produced from the balanced equation.

This is illustrated below:

Molar mass of NO2 = 14 + (16x2) = 46 g/mol

Mass of NO2 from the balanced equation = 3 x 46 = 138 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 1 x 18 = 18 g

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO from the balanced equation = 1 x 30 = 30 g

From the balanced equation above,

138 g of NO2 reacted with 18 g of H2O to produce 30 g of NO.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

138 g of NO2 reacted with 18 g of H2O.

Therefore, 50 g of NO2 will reacted to produce = (50 x 18) /138 = 6.52 g of H2O.

From the calculations made above,

Only 6.52 g out 15 g of H2O given was required to react completely with 50 g of NO2.

Therefore, NO2 is the limiting reactant and H2O is the excess reactant

Finally, we shall determine the mass of NO produced from the reaction.

In this case, the limiting reactant shall be used.

The limiting reactant is NO2 and the mass of NO produced can be obtained as follow:

From the balanced equation above,

138 g of NO2 reacted to produce 30 g of NO.

Therefore, 50 g of NO2 will react to produce = (50 x 30)/138 = 10.87 g of NO.

Therefore, 10.87 g of NO were obtained from the reaction.

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