if an excess of nitrogen is reacted with 3.41 g of hydrogen, how many grams of ammonia can be produced?
First, write out the equation:
N2+3H2 <=>2NH3
No of mol of H2->
3.41/2
=1.705 mol
Ratio of NH3 to H2 is 2:3,
Mol of NH3= (1.705/3)*2 =1.137 mol
Mass=no of mol*molar mass
=1.137*(14+3)
=19.3g
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Ah, nitrogen and hydrogen getting together for a little chemical dance! Well, to calculate the number of grams of ammonia that can be produced, we must first determine the limiting reactant. See, if there's an excess of nitrogen, then hydrogen is the limiting reactant because it will run out first. Once we figure that out, we can calculate the amount of ammonia that can be produced.
Now, I could go on and spout a bunch of stoichiometry equations, but seriously, who needs all that seriousness? Let's cut to the chase. Assuming we're dealing with an idealized world, where reactions are perfectly efficient, and all that fun stuff, the answer is... drumroll, please... it's approximately 9.68 grams of ammonia!
Remember, though, in real-life scenarios, reactions aren't always 100% efficient, so the actual yield might vary. But hey, we can still have some fun with the calculations, right?
To solve this problem, we need to determine the limiting reactant first. This will tell us which reactant will be completely consumed and determine the maximum amount of ammonia that can be produced.
Step 1: Write and balance the chemical equation.
N2 + 3H2 -> 2NH3
Step 2: Calculate the molar mass of nitrogen (N2).
Nitrogen (N2) = 14.007 g/mol * 2 = 28.014 g/mol
Step 3: Calculate the number of moles of nitrogen.
moles of nitrogen = mass of nitrogen / molar mass of nitrogen
moles of nitrogen = 3.41 g / 28.014 g/mol
Step 4: Calculate the molar mass of hydrogen (H2).
Hydrogen (H2) = 1.008 g/mol * 2 = 2.016 g/mol
Step 5: Calculate the number of moles of hydrogen.
moles of hydrogen = mass of hydrogen / molar mass of hydrogen
moles of hydrogen = 3.41 g / 2.016 g/mol
Step 6: Calculate the stoichiometric ratio between nitrogen and ammonia.
From the balanced equation, we see that 1 mole of nitrogen reacts with 2 moles of ammonia.
Step 7: Determine the limiting reactant.
The reactant that produces the lesser amount of product is the limiting reactant. Compare the moles of nitrogen and moles of hydrogen to determine the limiting reactant.
moles of nitrogen * (2 moles of ammonia / 1 mole of nitrogen) = number of moles of ammonia that can be produced from nitrogen
moles of hydrogen * (2 moles of ammonia / 3 moles of hydrogen) = number of moles of ammonia that can be produced from hydrogen
Whichever result is smaller will determine the limiting reactant.
Step 8: Calculate the grams of ammonia produced.
grams of ammonia produced = moles of ammonia * molar mass of ammonia
After following these steps, you should be able to determine the grams of ammonia that can be produced.
To determine the grams of ammonia produced, we need to first determine which reactant is limiting and then calculate the amount of ammonia based on the balanced chemical equation.
1. Write the balanced chemical equation:
N2 + 3H2 -> 2NH3
2. Calculate the molar masses of nitrogen (N2), hydrogen (H2), and ammonia (NH3):
- Nitrogen (N2) molar mass = 2 * 14.01 g/mol = 28.02 g/mol
- Hydrogen (H2) molar mass = 2 * 1.01 g/mol = 2.02 g/mol
- Ammonia (NH3) molar mass = 14.01 g/mol + 3 * 1.01 g/mol = 17.04 g/mol
3. Convert the mass of hydrogen (H2) to moles:
Moles of H2 = Mass (g) / Molar mass (g/mol)
Moles of H2 = 3.41 g / 2.02 g/mol = 1.6881 mol (rounded to four decimal places)
4. Use the stoichiometry of the balanced chemical equation to determine the moles of ammonia (NH3) produced:
According to the balanced equation, 3 moles of hydrogen (H2) react with 1 mole of ammonia (NH3). Therefore, the moles of NH3 produced will be equal to the moles of H2 used.
Moles of NH3 = Moles of H2 = 1.6881 mol (rounded to four decimal places)
5. Convert the moles of ammonia (NH3) to grams:
Grams of NH3 = Moles of NH3 * Molar mass of NH3
Grams of NH3 = 1.6881 mol * 17.04 g/mol = 28.76 g (rounded to two decimal places)
Therefore, with an excess of nitrogen, 3.41 g of hydrogen can produce approximately 28.76 grams of ammonia.