Show how you would calculate the % yield of hydrogen if 40.0 grams of magnesium react with an excess of nitric acid producing 1.70 grams of hydrogen gas.
Mg + 2 HNO3 ---> Mg(NO3)2 + H2
mols Mg = grams/atomic mass = ?
Using the coefficients in the balanced equation, convert mols Mg to mols H2. You can see it's a 1:1 ratio; therefore, mols Mg = mols H2 produced.
Then convert mols H2 to grams. g H2 = mols H2 x molar mass H2 = ? This is the theoretical yield (TY). The actual yield (AY) in the problem is listed as 1.70 g.
% yield = (AY/TY)*100 = >
Thanks @DrBob222
To calculate the percent yield of hydrogen gas in this reaction, you need to compare the amount of hydrogen actually obtained (1.70 grams) to the theoretical yield of hydrogen that could be obtained from the given amount of magnesium (40.0 grams).
Step 1: Calculate the moles of magnesium (Mg).
Molar mass of Mg = 24.31 g/mol
Moles of Mg = Mass of Mg / Molar mass of Mg
Moles of Mg = 40.0 g / 24.31 g/mol
Moles of Mg = 1.645 mol (rounded to three decimal places)
Step 2: Use the balanced equation to determine the stoichiometric ratio between magnesium and hydrogen gas.
From the balanced equation: 1 mol of Mg produces 1 mol of H2.
Therefore, 1.645 mol of Mg produces 1.645 mol of H2.
Step 3: Calculate the theoretical yield of hydrogen gas in grams.
Molar mass of H2 = 2.02 g/mol
Theoretical yield of H2 = Moles of H2 x Molar mass of H2
Theoretical yield of H2 = 1.645 mol x 2.02 g/mol
Theoretical yield of H2 = 3.325 g (rounded to three decimal places)
Step 4: Calculate the percent yield of hydrogen gas.
Percent Yield = (Actual yield / Theoretical yield) x 100
Percent Yield = (1.70 g / 3.325 g) x 100
Percent Yield ≈ 51.13%
Therefore, the percent yield of hydrogen gas in this reaction is approximately 51.13%.
To calculate the percentage yield of hydrogen in this reaction, we need to compare the actual yield of hydrogen with the theoretical yield of hydrogen.
Step 1: Determine the balanced chemical equation:
The balanced equation for the reaction is:
Mg + 2 HNO3 ---> Mg(NO3)2 + H2
Step 2: Calculate the molar mass:
- The molar mass of magnesium (Mg) is 24.31 g/mol.
- The molar mass of hydrogen (H2) is 2.02 g/mol.
Step 3: Calculate the moles of magnesium used:
We have 40.0 grams of magnesium, so to calculate the moles, we divide the mass by the molar mass:
40.0 g / 24.31 g/mol = 1.6468 mol Mg
Step 4: Determine the stoichiometric ratio:
From the balanced equation, we see that 1 mole of magnesium reacts with 1 mole of hydrogen. Therefore, the moles of hydrogen should be the same as the moles of magnesium.
Step 5: Calculate the theoretical yield of hydrogen:
The theoretical yield represents the maximum amount of hydrogen that can be produced in the reaction. Since the moles of hydrogen should be the same as the moles of magnesium, we can simply multiply the moles of magnesium by the molar mass of hydrogen to get the theoretical yield:
1.6468 mol Mg * 2.02 g/mol H2 = 3.326 g H2
Step 6: Calculate the percentage yield:
The percentage yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100%:
Percentage yield = (Actual yield / Theoretical yield) * 100%
Actual yield = 1.70 g
Theoretical yield = 3.326 g
Percentage yield = (1.70 g / 3.326 g) * 100% ≈ 51.18%
Therefore, the percentage yield of hydrogen in this reaction is approximately 51.18%.