Ammonia (NH3) reacts with oxygen to produce nitric oxide (NO) and water (see balanced equation below). How many grams of oxygen do you need to react with 21.4 g ammonia?
__4_NH3 + __5_O2 → __4_NO + _6__H2O
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
To determine the number of grams of oxygen required to react with 21.4 g of ammonia (NH3), we need to use the balanced chemical equation provided. The coefficients in the balanced equation represent the molar ratios between the reactants and products.
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.