Answer:
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.
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
Learn more: https://brainly.com/question/165414
