When a 3.80 g sample of magnesium nitride (MW 101g/mol) is reacted with 3.30 g of water, 3.60 g of MgO is formed. What is the percent yield of this reaction?
Mg3N2 + 3 H2O --> 2 NH3 + 3 MgO
79.1
To find the percent yield of a reaction, you need to compare the actual yield to the theoretical yield.
First, calculate the moles of magnesium nitride (Mg3N2) and water (H2O) used in the reaction:
Moles of Mg3N2 = Mass of Mg3N2 / Molar mass of Mg3N2
Moles of Mg3N2 = 3.80 g / 101 g/mol
Moles of H2O = Mass of H2O / Molar mass of H2O
Moles of H2O = 3.30 g / 18 g/mol
Next, using the balanced equation for the reaction, calculate the theoretical yield of MgO:
From the balanced equation, the mole ratio between Mg3N2 and MgO is 3:3. Therefore, the moles of MgO formed will be equal to the moles of Mg3N2 used.
Theoretical yield of MgO = Moles of Mg3N2
Now, calculate the actual yield of MgO:
Actual yield of MgO = 3.60 g
Finally, calculate the percent yield using the formula:
Percent yield = (Actual yield / Theoretical yield) × 100
Substitute the values:
Percent yield = (3.60 g / (3.80 g / 101 g/mol)) × 100
Calculate the value:
Percent yield = (3.60 g / 0.038 g) × 100
Percent yield = 9473.68
Therefore, the percent yield of this reaction is 9473.68%.
To calculate the percent yield of a reaction, we need to compare the actual yield (the amount of product obtained in the experiment) to the theoretical yield (the maximum amount of product that could be obtained).
First, we need to find the number of moles of magnesium nitride (Mg3N2) and water (H2O) used in the reaction. To do this, we divide the given masses by their respective molar masses:
Moles of Mg3N2 = Mass of Mg3N2 / Molar mass of Mg3N2
= 3.80 g / 101 g/mol
Moles of H2O = Mass of H2O / Molar mass of H2O
= 3.30 g / 18 g/mol
Next, we need to determine the limiting reactant, which is the reactant that will be completely consumed in the reaction, limiting the amount of product that can be formed. To do this, we compare the stoichiometric ratio of the reactants to the moles determined above.
From the balanced chemical equation, we can see that the stoichiometric ratio between Mg3N2 and H2O is 1:3. Therefore, we need 3 times as many moles of H2O as moles of Mg3N2.
If the moles of H2O calculated above is less than 3 times the moles of Mg3N2, then H2O is the limiting reactant. If not, Mg3N2 is the limiting reactant.
After identifying the limiting reactant, we need to calculate the theoretical yield of the product (MgO) using the stoichiometry of the balanced chemical equation. From the equation, we can see that the stoichiometric ratio between Mg3N2 and MgO is 1:3. Therefore, the moles of MgO formed will be three times the moles of Mg3N2 (or the moles of H2O, if it is the limiting reactant).
The theoretical yield of MgO can be calculated as follows:
Theoretical yield of MgO = 3 x Moles of limiting reactant (Mg3N2 or H2O) x Molar mass of MgO
Finally, we can calculate the percent yield using the actual yield and the theoretical yield:
Percent yield = (Actual yield / Theoretical yield) x 100%
Given that the mass of MgO formed is 3.60 g, we can now proceed to calculate the percent yield.