9.70 grams of nitrogen gas are allowed to react with 5.44 grams of oxygen gas.
What is the maximum amount of nitrogen monoxide that can be formed?
What amount of the excess reagent remains after the reaction is complete?
since N2 + O2 = 2NO,
each mole of N2 requires 1 mole of O2.
convert your grams to moles, and the molecule with the fewer moles is the limiting reagent.
To find the maximum amount of nitrogen monoxide that can be formed, we need to determine which reactant is the limiting reactant.
Step 1: Convert grams to moles for each reactant:
- Nitrogen gas (N2): 9.70 grams
- Oxygen gas (O2): 5.44 grams
Using the molar masses:
- Molar mass of N2 = 28.02 grams/mol
- Molar mass of O2 = 32.00 grams/mol
Moles of N2 = 9.70 grams / 28.02 grams/mol = 0.346 mol
Moles of O2 = 5.44 grams / 32.00 grams/mol = 0.170 mol
Step 2: Write the balanced chemical equation to determine the stoichiometry of the reactants and products:
N2 + O2 -> 2NO
According to the balanced equation, the mole ratio between N2 and NO is 1:2. This means that for every 1 mole of N2, we can form 2 moles of NO.
Step 3: Compare the mole ratios between the reactants:
From the balanced equation, we see that the mole ratio between N2 and O2 is 1:1. This means that both reactants have the same number of moles. Therefore, N2 is not in excess and O2 is not in excess.
Step 4: Determine the limiting reactant:
Since both reactants are in stoichiometric proportions, neither is in excess, and the reaction will consume all the reactants. Therefore, the limiting reactant is both N2 and O2.
Step 5: Calculate the maximum amount of nitrogen monoxide produced:
From the stoichiometry of the balanced chemical equation, we know that for every mole of N2, 2 moles of NO are produced. Therefore, the maximum amount of NO produced can be calculated using the mole ratio:
Moles of NO = Moles of N2 x (2 moles of NO / 1 mole of N2)
Moles of NO = 0.346 mol x (2 moles of NO / 1 mole of N2) = 0.692 mol
Step 6: Convert moles to grams for nitrogen monoxide:
Using the molar mass of NO (30.01 grams/mol):
Grams of NO = Moles of NO x Molar mass of NO
Grams of NO = 0.692 mol x 30.01 grams/mol = 20.77 grams
Therefore, the maximum amount of nitrogen monoxide that can be formed is 20.77 grams.
To find the amount of excess reagent that remains after the reaction is complete, we can subtract the moles of the limiting reactant used from the original moles of the excess reactant.
Moles of excess O2 remaining = Moles of O2 - Moles of NO used
Moles of excess O2 remaining = 0.170 mol - (0.692 mol x (1 mole of O2 / 2 moles of NO))
Moles of excess O2 remaining = 0.170 mol - 0.346 mol x (1 mole of O2 / 2 moles of NO)
Moles of excess O2 remaining = 0.170 mol - 0.173 mol = -0.003 mol
The negative value implies that there is no excess O2 remaining after the reaction is complete.
Therefore, there is no excess reagent remaining after the reaction is complete.
To determine the maximum amount of nitrogen monoxide (NO) that can be formed and the amount of the excess reagent remaining, we need to use stoichiometry and the concept of limiting and excess reagents.
1. Write the balanced chemical equation for the reaction between nitrogen gas (N2) and oxygen gas (O2) to form nitrogen monoxide (NO):
N2 + O2 → 2NO
2. Calculate the number of moles for each reactant:
Moles of N2 = mass of N2 / molar mass of N2
= 9.70 g / 28.01 g/mol
≈ 0.3469 mol
Moles of O2 = mass of O2 / molar mass of O2
= 5.44 g / 32.00 g/mol
≈ 0.1700 mol
3. Determine the limiting reagent by comparing the mole ratio between N2 and O2 from the balanced equation (1 mole N2:1 mole O2:2 moles NO):
N2:O2
0.3469 mol:0.1700 mol
The ratio tells us that we have more N2 than O2, so O2 is the limiting reagent.
4. Calculate the maximum amount of NO that can be formed using the mole ratio between O2 and NO:
Moles of NO = moles of O2 * (2 moles NO / 1 mole O2)
= 0.1700 mol * (2 moles NO / 1 mole O2)
= 0.3400 mol
To convert this to grams, we use the molar mass of NO:
Mass of NO = moles of NO * molar mass of NO
= 0.3400 mol * 30.01 g/mol
≈ 10.20 g
Therefore, the maximum amount of nitrogen monoxide (NO) that can be formed is approximately 10.20 grams.
5. To find the amount of excess reagent remaining, we need to determine the amount of O2 that reacted. This can be done by subtracting the moles of NO produced from the moles of O2 initially present:
Moles of O2 remaining = Moles of O2 initial - (moles of NO * (1 mole O2 / 2 moles NO))
= 0.1700 mol - (0.3400 mol * (1 mole O2 / 2 moles NO))
= 0.1700 mol - 0.1700 mol
= 0 mol
Since the entire amount of O2 reacted, there is no excess O2 remaining.
In conclusion, the maximum amount of nitrogen monoxide (NO) that can be formed is approximately 10.20 grams, and there is no excess reagent (O2) remaining after the reaction is complete.