72g of Mg reacted with 32g of oxygen to form 48g of Mg. So %yield of Mgo
To calculate the percent yield of MgO (magnesium oxide), we need to compare the actual yield (the amount of MgO formed) with the theoretical yield (the amount of MgO that would be obtained if the reaction went to completion without any loss).
First, let's find the theoretical yield of MgO.
1. Determine the molar mass of Mg and O:
- Mg: 24.31 g/mol
- O: 16.00 g/mol
2. Convert the masses of Mg and O into moles:
- Moles of Mg = mass of Mg / molar mass of Mg
= 72 g / 24.31 g/mol
= 2.96 mol (rounded to two decimal places)
- Moles of O = mass of O / molar mass of O
= 32 g / 16.00 g/mol
= 2 mol
3. Determine the stoichiometry of the reaction. From the balanced chemical equation, we know that 2 moles of Mg react with 1 mole of O to produce 2 moles of MgO.
4. Calculate the theoretical yield of MgO:
- The limiting reactant is the one that produces the least amount of product. In this case, Mg is the limiting reactant since it has a stoichiometric ratio of 2:2, while O has a stoichiometric ratio of 2:1.
- For every 2 moles of Mg, 2 moles of MgO are formed. Therefore, with 2.96 moles of Mg (the limiting reactant), the theoretical yield of MgO is:
Theoretical yield = 2.96 mol Mg * (2 mol MgO / 2 mol Mg)
= 2.96 mol
Now, let's calculate the percent yield of MgO.
Percent yield = (actual yield / theoretical yield) * 100
Given that the actual yield of MgO is 48g, and the theoretical yield is 2.96 mol, we need to convert grams to moles for the actual yield:
Actual yield of MgO = 48 g / molar mass of MgO
To find the molar mass of MgO:
- Mg: 24.31 g/mol
- O: 16.00 g/mol
Molar mass of MgO = molar mass of Mg + molar mass of O
= 24.31 g/mol + 16.00 g/mol
= 40.31 g/mol
Converting grams to moles:
Actual yield of MgO = 48 g / 40.31 g/mol
Now that we have the actual yield in moles, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100
= (actual yield in moles / theoretical yield) * 100
Substituting the values:
Percent yield = (48 g / 40.31 g/mol) / 2.96 mol) * 100
= (1.19 / 2.96) * 100
= 40.20% (rounded to two decimal places)
Therefore, the percent yield of MgO is approximately 40.20%.