1.29g H2 is allowed to react with 10.3g N2 , producing 1.86g NH3.
What is the theoretical yield for this reaction under the given conditions?
What is the percent yield for this reaction under the given conditions?
I have been stucks for days now.
There is a long way and a short way to work these limiting reagent problems. I use the long way.
1. Write and balance the equation.
N2 + 3H2 ==> 2NH3
2a. Convert 1.29 g H2 to moles. moles = grams/molar mass = 1.29/2 = 0.645 moles H2.
2b. Same for N2. moles N2 = 10.3/28 = 0.368.
3a. Using the coefficients in the balanced equation, convert moles H2 to moles of the product, in this case, NH3.
0.645 moles H2 x (2 mol NH3/3 mol H2) = 0.43 x (2/3) = 0.430 moles NH3.
3b. Same for N2.
0.368 N2 x (2 moles NH3/1 mol N2) = 0.368 x (2/1) = 0.736 mole NH3.
3c. You see there are two answers from 3a and 3b; obviously only one can be correct. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. (You should understand why this is true.)
4. Using the smaller value, convert 0.430 moles NH3 to grams. This is the theoretical yield of NH3 in grams. (The 0.430 was the theoretical yield in moles.). 0.430 moles NH3 x (17 g NH3/1 mole NH3) = 7.31 grams NH3 produced. That is the maximum amount that can be produced. What happens to the extra N2. It just sits there. Enough is supplied to react with all of the hydrogen and the excess remains unused.
Then %yield = (actual yield/theoretical yield)*100 = (1.86/7.31)*100 = ?
Print this out. It will work all of your limiting reagent problems.
To find the theoretical yield, we need to determine the limiting reactant. The limiting reactant is the reactant that will be completely consumed, thus limiting the amount of product that can be formed.
1. Calculate the number of moles for each reactant:
- H2: 1.29 g / molar mass of H2 (2 g/mol) = 0.645 mol H2
- N2: 10.3 g / molar mass of N2 (28 g/mol) = 0.368 mol N2
2. Use the balanced equation:
N2 + 3H2 -> 2NH3
From the balanced equation, we can see that 1 mol of N2 reacts with 3 mol of H2 to produce 2 mol of NH3.
3. Determine the limiting reactant:
We can now compare the ratios of the moles of reactants to the stoichiometric coefficients in the balanced equation.
- For N2: 0.368 mol N2 × (3 mol H2 / 1 mol N2) = 1.104 mol H2
- For H2: 0.645 mol H2
Since the stoichiometric ratio is 3 mol H2 to 1 mol N2, we can see that N2 is the limiting reactant as it requires 3 times more H2 than what is available.
4. Calculate the theoretical yield:
From the balanced equation, we know that 2 mol of NH3 are produced from 1 mol of N2. Therefore, the theoretical yield of NH3 is:
0.368 mol N2 × (2 mol NH3 / 1 mol N2) × molar mass of NH3 (17 g/mol) = 12.448 g NH3
To find the percent yield, we compare the actual yield (given as 1.86 g NH3) to the theoretical yield calculated above.
5. Calculate the percent yield:
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (1.86 g / 12.448 g) × 100 = 14.93%
Therefore, the theoretical yield for this reaction is 12.448 g NH3, and the percent yield is 14.93%.
To determine the theoretical yield and percent yield of a reaction, we need to follow a few steps.
Step 1: Write and balance the chemical equation for the reaction.
The balanced chemical equation for the reaction between hydrogen gas (H2) and nitrogen gas (N2) to produce ammonia (NH3) is as follows:
3H2 + N2 -> 2NH3
Step 2: Calculate the molar masses of the reactants and product.
- The molar mass of H2 is 2 grams/mol (2 g/mol).
- The molar mass of N2 is 28 grams/mol (28 g/mol).
- The molar mass of NH3 is 17 grams/mol (17 g/mol).
Step 3: Convert the given masses of reactants to moles.
- For H2: 1.29 g of H2 * (1 mol H2/2 g H2) = 0.645 mol H2
- For N2: 10.3 g of N2 * (1 mol N2/28 g N2) = 0.368 mol N2
Step 4: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio from the balanced equation. The reactant that produces fewer moles of product is the limiting reactant.
From the balanced equation, the stoichiometric ratio is 3 moles of H2 to 1 mole of N2.
- For H2: 0.645 mol H2 * (1 mol N2 / 3 mol H2) = 0.215 mol N2
Since we have 0.215 mol N2 and the actual amount of N2 is 0.368 mol, N2 is the limiting reactant.
Step 5: Determine the theoretical yield.
The theoretical yield is the maximum amount of product that can be formed from the limiting reactant.
From the balanced equation, the stoichiometric ratio is 3 moles of H2 to 2 moles of NH3.
- For N2: 0.368 mol N2 * (2 mol NH3 / 1 mol N2) = 0.736 mol NH3
The theoretical yield of NH3 is 0.736 mol.
Step 6: Calculate the percent yield.
The percent yield is the ratio of the actual yield (given in the question) to the theoretical yield, multiplied by 100.
- Actual yield = 1.86 g of NH3
- Theoretical yield = 0.736 mol of NH3 * molar mass of NH3 (17 g/mol)
Percent yield = (actual yield / theoretical yield) * 100
Let's calculate the percent yield:
Percent yield = (1.86 g / (0.736 mol * 17 g/mol)) * 100 = 166.30%
Theoretical yield: 0.736 mol of NH3
Percent yield: 166.30%
Note: The percent yield can exceed 100% if there are errors in the experimental procedure or if impurities are present in the reactants or products.