a)What is the theoretical yield for this reaction under the given conditions?
b)What is the percent yield for this reaction under the given conditions?
1.41gH2 is allowed to react with 10.4gN2,producing 2.46gNH3
To determine the theoretical yield and percent yield for a reaction, you will need to use stoichiometry and convert the given masses of reactants and products to moles.
a) The theoretical yield is the maximum amount of product that can be obtained from a reaction, assuming complete conversion of reactants to products. In this case, the balanced equation for the reaction is:
N2 + 3H2 → 2NH3
According to the equation, one mole of N2 reacts with three moles of H2 to produce two moles of NH3. First, you need to convert the given masses of H2 and N2 to moles by dividing the mass by the molar mass:
moles of H2 = 1.41g H2 / molar mass of H2
moles of N2 = 10.4g N2 / molar mass of N2
Once you have determined the number of moles for each reactant, you can compare the mole ratio from the balanced equation to calculate the moles of NH3 that should be produced. In this case, since the reaction produces 2.46g of NH3, you can convert this mass to moles using the molar mass of NH3:
moles of NH3 = 2.46g NH3 / molar mass of NH3
The calculated moles of NH3 will be the theoretical yield of the reaction.
b) The percent yield is a measure of the efficiency of a reaction and is given by the formula:
Percent yield = (actual yield / theoretical yield) * 100
To calculate the percent yield, divide the actual yield (given in the problem) by the theoretical yield (calculated in part a). Finally, multiply by 100 to express the result as a percentage.
To find the theoretical yield and percent yield, we need to determine the stoichiometry of the reaction. According to the balanced chemical equation:
3H2 + N2 → 2NH3
Given:
Mass of H2 = 1.41g
Mass of N2 = 10.4g
Mass of NH3 = 2.46g
a) Theoretical yield:
To calculate the theoretical yield, we need to determine the limiting reagent. We can do this by comparing the amount of product that can be formed from each reactant assuming complete consumption.
First, let's find the number of moles of each reactant:
Molar mass of H2 = 2.02 g/mol
Molar mass of N2 = 28.02 g/mol
Molar mass of NH3 = 17.03 g/mol
Number of moles of H2 = Mass of H2 / Molar mass of H2
= 1.41g / 2.02 g/mol
= 0.699 moles
Number of moles of N2 = Mass of N2 / Molar mass of N2
= 10.4g / 28.02 g/mol
= 0.371 moles
Using the stoichiometry of the balanced equation, we can determine the amount of NH3 that would be formed assuming all reactants are consumed:
From the reaction: 3H2 + N2 → 2NH3
Given: Number of moles of H2 = 0.699 moles
Number of moles of N2 = 0.371 moles
The ratio of moles of H2 to moles of NH3 is 3:2. Therefore, the limiting reagent is H2 because it produces fewer moles of NH3 compared to N2.
To find the theoretical yield, we'll use the ratio of moles of H2 to moles of NH3:
Moles of NH3 = (Number of moles of H2) x (2 moles of NH3 / 3 moles of H2)
= 0.699 moles x (2/3)
= 0.466 moles
Molar mass of NH3 = 17.03 g/mol
Theoretical yield = Moles of NH3 x Molar mass of NH3
= 0.466 moles x 17.03 g/mol
= 7.94 g
Therefore, the theoretical yield of NH3 is 7.94 grams.
b) Percent yield:
The percent yield is calculated by comparing the actual yield to the theoretical yield and expressing it as a percentage:
Percent yield = (Actual yield / Theoretical yield) x 100
Given: Actual yield = 2.46 g (as mentioned in the question)
Percent yield = (2.46 g / 7.94 g) x 100
= 31.01%
Therefore, the percent yield of the reaction is 31.01%.