What mass of nitrogen monoxide can be made by the reaction of 8.00 g ammonia and 17.0 g oxygen?
correct answer is 12.8 ± 0.1 (g)
I keep getting 14.1 g
and that the limiting reagent is NH3
What am I doing wrong?
I suspect you didn't write and balance the equation. I worked it and obtained 12.8g and O2 is the limiting reagent.
Equation is
4NH3 + 5O2 ==> 4NO + 6H2O
yep that was defiantly the problem.
thank you
To determine the correct mass of nitrogen monoxide produced and identify the limiting reagent, we need to follow a step-by-step process.
Step 1: Write the balanced equation for the reaction between ammonia (NH3) and oxygen (O2) to form nitrogen monoxide (NO):
4 NH3 + 5 O2 -> 4 NO + 6 H2O
Step 2: Calculate the molar masses of the reactants and the product:
NH3: 1 nitrogen (N) atom + 3 hydrogen (H) atoms = 14.01 g/mol
O2: 2 oxygen (O) atoms = 32.00 g/mol
NO: 1 nitrogen (N) atom + 1 oxygen (O) atom = 30.01 g/mol
Step 3: Convert the given masses of ammonia (NH3) and oxygen (O2) to moles:
NH3: 8.00 g NH3 * (1 mol NH3 / 14.01 g NH3) ≈ 0.571 mol NH3
O2: 17.0 g O2 * (1 mol O2 / 32.00 g O2) ≈ 0.531 mol O2
Step 4: Determine the mole ratio between NH3 and NO from the balanced equation:
From the balanced equation, 4 moles of NH3 react to form 4 moles of NO.
Step 5: Calculate the maximum theoretical yield of NO, assuming the reaction goes to completion:
0.571 mol NH3 * (4 mol NO / 4 mol NH3) = 0.571 mol NO
Step 6: Determine the mole ratio between O2 and NO from the balanced equation:
From the balanced equation, 5 moles of O2 react to form 4 moles of NO.
Step 7: Calculate the maximum theoretical yield of NO, assuming the reaction goes to completion:
0.531 mol O2 * (4 mol NO / 5 mol O2) = 0.425 mol NO
Step 8: Identify the limiting reagent by comparing the moles of NH3 and O2:
Since we calculated that 0.571 mol NO can be formed from NH3, and only 0.425 mol NO can be formed from O2, the limiting reagent is O2.
Step 9: Calculate the mass of NO produced from the limiting reagent:
0.425 mol NO * (30.01 g NO / 1 mol NO) ≈ 12.8 ± 0.1 g NO
Therefore, the correct mass of nitrogen monoxide that can be made is approximately 12.8 ± 0.1 g, not 14.1 g.
To determine what you might be doing wrong, let's go through the steps to solve this problem.
1. Write the balanced equation for the reaction: 4NH3 + 5O2 -> 4NO + 6H2O.
2. Calculate the molar mass of each substance:
- Molar mass of ammonia (NH3) = 14.01 g/mol
- Molar mass of oxygen (O2) = 32.00 g/mol
- Molar mass of nitrogen monoxide (NO) = 30.01 g/mol
3. Convert the given masses (8.00 g of ammonia and 17.0 g of oxygen) into moles:
- Moles of ammonia = 8.00 g / 14.01 g/mol = 0.571 mol
- Moles of oxygen = 17.0 g / 32.00 g/mol = 0.531 mol
4. Determine the limiting reagent by comparing the mole ratios of the reactants in the balanced equation:
- According to the balanced equation, 4 moles of NH3 react with 5 moles of O2 to produce 4 moles of NO.
- Since the ratio of NH3 to O2 in the given amounts is approximately 4:5, we can conclude that NH3 is the limiting reagent.
5. Calculate the mass of nitrogen monoxide produced using the limiting reagent:
- To do this, we need to first determine the moles of nitrogen monoxide produced. According to the balanced equation, 4 moles of NH3 produce 4 moles of NO.
- Since NH3 is the limiting reagent, the moles of NO produced will be equal to the moles of NH3.
- Moles of NO produced = 0.571 mol
6. Finally, convert moles of NO to grams:
- Mass of NO = moles of NO × molar mass of NO
- Mass of NO = 0.571 mol × 30.01 g/mol = 17.14 g
- Rounding to the correct number of significant figures, the mass of NO is 17.1 g.
Based on these calculations, it seems that the correct answer is indeed 17.1 g, not 14.1 g as you mentioned. So it appears there might have been a mistake in your calculations. I would suggest re-checking your work and ensuring that you have followed the steps correctly.