mols NH3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols NH3 to mols NO.
That's ?mols NH3 x (4 mols NO/4 mols NH3) = ?mols NH3 x 1/1 = ?
Then grams NO = mols NO x molar mass NO
__4_NH3 + __5_O2 → __4_NO + _6__H2O
Using the coefficients in the balanced equation, convert mols NH3 to mols NO.
That's ?mols NH3 x (4 mols NO/4 mols NH3) = ?mols NH3 x 1/1 = ?
Then grams NO = mols NO x molar mass NO
In this case, the balanced equation is:
4 NH3 + 5 O2 → 4 NO + 6 H2O
From the balanced equation, we can see that 5 moles of oxygen (O2) are required to react with 4 moles of ammonia (NH3).
To find the number of moles of ammonia, we can divide the given mass (21.4 g) by the molar mass of ammonia (17.03 g/mol).
Number of moles of ammonia = mass of ammonia / molar mass of ammonia
Number of moles of ammonia = 21.4 g / 17.03 g/mol ≈ 1.257 mol
Now, using the molar ratio from the balanced equation, we can determine the number of moles of oxygen required.
Number of moles of oxygen = (number of moles of ammonia) × (coefficient of O2 / coefficient of NH3)
Number of moles of oxygen = 1.257 mol × (5 / 4) = 1.571 mol
Finally, to find the mass of oxygen, we can multiply the number of moles of oxygen by the molar mass of oxygen (32.00 g/mol).
Mass of oxygen = number of moles of oxygen × molar mass of oxygen
Mass of oxygen = 1.571 mol × 32.00 g/mol ≈ 50.272 g
Therefore, approximately 50.272 grams of oxygen are required to react with 21.4 grams of ammonia.