The balanced chemical equation for the combustion of ammonia is:

4 NH3 + 5 O2 → 4 NO2 + 6 H2O

According to the stoichiometry of this equation, 4 moles of NH3 produce 4 moles of NO2.

To calculate the amount of NO2 produced, we need to convert the mass of NH3 to moles, then use the mole ratio from the balanced equation.

First, we find the molar mass of NH3:
Nitrogen (N) has an atomic mass of 14.01 g/mol.
Hydrogen (H) has an atomic mass of 1.01 g/mol.

Molar mass of NH3 = 14.01 g/mol + (3 * 1.01 g/mol)
= 14.01 g/mol + 3.03 g/mol
= 17.04 g/mol.

Next, we convert the mass of NH3 to moles using its molar mass:
Moles of NH3 = mass (g) / molar mass (g/mol)
= 100 g / 17.04 g/mol
≈ 5.86 mol.

Since the ratio between NH3 and NO2 in the balanced equation is 4:4, we can say that the moles of NO2 produced will also be 5.86 mol.

Finally, we convert the moles of NO2 to grams using the molar mass of NO2. Nitrogen (N) has an atomic mass of 14.01 g/mol, and Oxygen (O) has an atomic mass of 16.00 g/mol.

Molar mass of NO2 = (14.01 g/mol) + (2 * 16.00 g/mol)
= 14.01 g/mol + 32.00 g/mol
= 46.01 g/mol.

Mass of NO2 = moles of NO2 * molar mass of NO2
= 5.86 mol * 46.01 g/mol
≈ 271 g.

Therefore, approximately 271 grams of nitrogen dioxide (NO2) would be produced when 100 grams of ammonia (NH3) react with oxygen (O2) through complete combustion.

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