How many grams of nitrogen are required to react with 2.79g of hydrogen to produce ammonia?
A) 25.8 g B) 13.0 g C) 78.2 g D) 38.7 g E) 77.4 g
I get the answer as 38.7 but it is meant to be B
I did 2.79g x 3molH2/2.016g/mol x 1molN2/3molH2 x 28.02g/mol/1moleN2
Whats the equation to form NH2?
N2 +3H2 -----> 2NH3
So, you need 3 moles of H2 gas for 1 mole of N2:
How many moles of hydrogen are present in 2.79 g of Hydrogen?
2.79 g of H2*(1 mole/2.016g)=1.384 moles of H2
1.384 moles of H2 requires how many moles of N2?
1.384 moles of H2*(1 mole of N2/3 moles of H2)=0.4613 moles of N2
0.4613 moles of N2*(28.014g/mole)=12.9 g of N2=13.0g of N2.
I do it in steps when explaining. Also, it is better for accounting reasons in order to find errors. But you could do it all in one step, and you should still get the same answer:
2.79 g of H2*(1 mol/2.016g)*(1 mol N2/3 mol of H2)*(28.014g/mole)= 13.0g of N2
Well, it seems like you're trying to balance the equation and calculate the amount of nitrogen needed in grams. But let me see if I can bring a little humor to your question!
Why did the hydrogen and nitrogen decide to make ammonia together? They heard it was a-moo-sing! (Get it? Am-oo-sing? Ammonia? Okay, maybe I'll stick to answering the question now.)
So, let's work out the math. You started off on the right track with the conversions, but it looks like you made a small calculation error along the way. Let's try it step by step:
2.79g H2 x (1 mol H2 / 2.016g) x (1 mol N2 / 3 mol H2) x (28.02g / 1 mol N2)
After doing the math, you should end up with approximately 13.01g of nitrogen. So, the correct answer is indeed B) 13.0g!
Don't worry, chemistry can be a bit tricky sometimes. But as long as you're having fun with it, you'll learn to harness the power of chemical reactions and laughs!
To find the correct answer, let's go through the calculation step-by-step.
Given:
Mass of hydrogen (H2) = 2.79 g
We need to determine the mass of nitrogen (N2) required to produce ammonia (NH3) by reacting with the given mass of hydrogen.
The balanced chemical equation for the reaction is:
N2 + 3H2 -> 2NH3
Step 1: Convert the given mass of hydrogen (H2) to moles.
molar mass of H2 = 2.016 g/mol
moles of H2 = mass of H2 / molar mass of H2
= 2.79 g / 2.016 g/mol
= 1.384 moles
Step 2: Use the stoichiometry from the balanced equation to determine the moles of nitrogen (N2) required.
From the balanced equation, we see that 3 moles of H2 react with 1 mole of N2.
moles of N2 = moles of H2 / 3
= 1.384 moles / 3
≈ 0.461 moles
Step 3: Convert the moles of nitrogen (N2) to grams.
molar mass of N2 = 28.02 g/mol
mass of N2 = moles of N2 * molar mass of N2
= 0.461 moles * 28.02 g/mol
≈ 12.9 g
Therefore, the correct answer is B) 13.0 g.
Your calculation is almost correct, but you made a small error in your calculation. Let's go through the steps again to find the correct answer:
To solve this problem, we need to use the balanced chemical equation for the reaction between nitrogen and hydrogen to produce ammonia:
N2 + 3H2 → 2NH3
1. Start with the given mass of hydrogen: 2.79g H2.
2. Convert the mass of hydrogen to moles using its molar mass (2.016g/mol) since we have the molar ratio between nitrogen and hydrogen in the equation.
2.79g H2 x (1 mol H2 / 2.016g H2) = 1.38 mol H2
3. Use the molar ratio from the balanced equation to find the moles of nitrogen needed. From the equation, we see that 3 moles of hydrogen react with 1 mole of nitrogen.
1.38 mol H2 x (1 mol N2 / 3 mol H2) = 0.46 mol N2
4. Finally, convert the moles of nitrogen to grams using its molar mass (28.02g/mol).
0.46 mol N2 x (28.02g N2 / 1 mol N2) = 12.9g N2, which is approximately 13.0g N2.
Therefore, the correct answer is B) 13.0g of nitrogen.