If 26.4 grams of nitrogen and 30.1 grams of oxygen combine to form nitrogen monoxide, how many grams of nitrogen monoxide must form? I am confused on how to do the math.
Well, calculating the number of grams of nitrogen monoxide formed involves a little bit of chemistry and math. So, let's break it down!
First, we need to determine the molecular weights of nitrogen (N), oxygen (O), and nitrogen monoxide (NO). The atomic masses for nitrogen and oxygen are approximately 14.01 g/mol and 16.00 g/mol, respectively. Adding them up gives us the molecular weight of nitrogen monoxide as approximately 30.01 g/mol.
Now, let's calculate the number of moles for nitrogen and oxygen by dividing their given masses by their respective atomic masses.
For nitrogen: 26.4 g ÷ 14.01 g/mol = approximately 1.8801 mol
For oxygen: 30.1 g ÷ 16.00 g/mol = approximately 1.8813 mol
Since the balanced chemical equation for the formation of nitrogen monoxide is:
2N2 + O2 → 2NO
We can see that for every 2 moles of nitrogen, we need 1 mole of oxygen to produce 2 moles of nitrogen monoxide.
Comparing the number of moles we have for nitrogen and oxygen, we can see that they are pretty much equal (1.8801 mol vs. 1.8813 mol). This means that we have the correct stoichiometric ratio of nitrogen to oxygen.
So, to determine the number of moles of nitrogen monoxide formed, we can use the stoichiometric ratio of 2N2 to 2NO. In this case, we have 1.8801 moles of nitrogen, which implies the formation of 1.8801 moles of nitrogen monoxide.
Finally, we multiply the moles of nitrogen monoxide by its molecular weight to find the mass in grams:
1.8801 mol × 30.01 g/mol ≈ 56.43 grams
Therefore, approximately 56.43 grams of nitrogen monoxide must form.
To find the amount of nitrogen monoxide (NO) that must form, you need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
To do this, you need to calculate the amount of nitrogen monoxide (NO) that can be formed from both nitrogen (N2) and oxygen (O2).
1. Start by calculating the number of moles for each reactant using their respective molar masses. The molar mass of nitrogen (N2) is 28.02 grams/mol, and the molar mass of oxygen (O2) is 32.00 grams/mol.
- Moles of nitrogen (N2): 26.4 g / 28.02 g/mol = 0.941 mol
- Moles of oxygen (O2): 30.1 g / 32.00 g/mol = 0.940 mol
2. Write the balanced equation for the reaction between nitrogen and oxygen to form nitrogen monoxide:
N2 + O2 -> 2NO
3. Next, find the molar ratio between nitrogen monoxide (NO) and each reactant from the balanced equation:
- Molar ratio between N2 and NO: 1 mol N2 : 2 mol NO
- Molar ratio between O2 and NO: 1 mol O2 : 2 mol NO
4. Now, determine the limiting reactant by comparing the number of moles of each reactant to the molar ratio with nitrogen monoxide (NO). The reactant that yields less nitrogen monoxide (NO) is the limiting reactant.
- For nitrogen: 0.941 mol N2 x (2 mol NO / 1 mol N2) = 1.882 mol NO
- For oxygen: 0.940 mol O2 x (2 mol NO / 1 mol O2) = 1.880 mol NO
5. Since oxygen (O2) yields less nitrogen monoxide (NO), it is the limiting reactant.
6. Finally, use the molar ratio between oxygen and nitrogen monoxide to calculate the mass of nitrogen monoxide that must form.
- Mass of NO: 0.940 mol NO x (30.01 g/mol) = 28.22 grams of NO
Therefore, 28.22 grams of nitrogen monoxide must form.