Nitric oxide, NO, is made from the oxidation of NH3, and the reaction is represented by the equation: 4NH3 + 5O2 → 4NO + 6H2O What mass of O2 would be required to react completely with 6.87 g of NH3?

Mass of oxygen required for the reaction is 16.16 g.

The equation of the reaction is;

4NH3 + 5O2 → 4NO + 6H2O

Since Number of moles = Mass/molar mass

Molar mass of ammonia = 17 g mol

Number of moles of NH3 reacted =  6.87 g/17 g mol = 0.404 moles

From the reaction equation;

4 moles of ammonia reacts with 5 moles of oxygen

0.404 moles reacts with 0.404 moles × 5 moles/4 moles  = 0.505 moles

Mass of oxygen = 0.505 moles × 32 g/mol = 16.16 g

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16.16g of O2.

Explanation:

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

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the mass of NH3 and O2 that reacted from the balanced equation. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160g.

From the balanced equation above,

68g of NH3 reacted with 160g of O2.

Now, we can calculate the mass of O2 that will be required to react completely with 6.87 g of NH3. This is illustrated below:

From the balanced equation above,

68g of NH3 reacted with 160g of O2

Therefore, 6.87g of NH3 will react with = (6.87 x 160)/68 = 16.16g of O2.

Therefore, 16.16g of O2 is needed for the reaction.