Magnesium hydroxide is produced by reacting magnesium oxide with excess water. If the reaction has an expected yield of 81%, how much magnesium oxide should be reacted to produce 98.0 g of magnesium hydroxide?
well, you could (at perfect yield), gotten 98.0/.81=121grams, or moles Mg(OH)2 121/58.3=2.07 moles
MgO+H2O>>>Mg(OH)2. From the balanced equation, you needed then 2.07 moles MgO, or 2.07*formulamassMgO grams
To determine the amount of magnesium oxide required to produce 98.0 g of magnesium hydroxide, we need to calculate the theoretical yield of the reaction.
The expected yield of 81% means that only 81% of the product is obtained compared to the maximum possible yield. Therefore, the theoretical yield can be calculated by dividing the desired amount of magnesium hydroxide by the percentage yield:
Theoretical yield = (Desired amount / Percentage yield)
Theoretical yield = (98.0 g / 81%)
Theoretical yield = 120.99 g (rounded to 3 decimal places)
Since the reaction is stoichiometric, the molar ratio between magnesium oxide (MgO) and magnesium hydroxide (Mg(OH)2) is 1:1. Therefore, the number of moles of magnesium oxide required is equal to the number of moles of magnesium hydroxide.
Now, we can calculate the number of moles of magnesium oxide using its molar mass:
Molar mass of MgO = 24.31 g/mol + 16.00 g/mol = 40.31 g/mol
Moles of MgO = (Theoretical yield / Molar mass of MgO)
Moles of MgO = (120.99 g / 40.31 g/mol)
Moles of MgO = 3.002 mol (rounded to 3 decimal places)
Since the molar ratio between MgO and Mg(OH)2 is 1:1, the number of moles of magnesium oxide required is also the number of moles of magnesium hydroxide produced.
Therefore, 3.002 moles or 120.99 g of magnesium oxide should be reacted to produce 98.0 g of magnesium hydroxide.
To solve this problem, we need to determine the amount of magnesium oxide needed to produce 98.0 g of magnesium hydroxide using the expected yield of 81%.
First, we need to understand the stoichiometry of the reaction. The balanced equation for the reaction of magnesium oxide (MgO) with water (H2O) to produce magnesium hydroxide (Mg(OH)2) is:
MgO + H2O -> Mg(OH)2
According to the balanced equation, 1 mole of MgO reacts with 1 mole of H2O to produce 1 mole of Mg(OH)2.
Next, we calculate the molar mass of Mg(OH)2. The molar mass of magnesium (Mg) is 24.31 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol. The molar mass of hydrogen (H) is 1.01 g/mol. Therefore, the molar mass of Mg(OH)2 is:
(24.31) + 2*(1.01) + 2*(16.00) = 58.33 g/mol
Now, we can calculate the number of moles of Mg(OH)2 needed to produce 98.0 g using the equation:
moles = mass / molar mass
moles = 98.0 g / 58.33 g/mol ≈ 1.68 mol
Since the expected yield is 81%, we need to adjust the number of moles accordingly:
actual moles = expected yield * moles
actual moles = 0.81 * 1.68 mol ≈ 1.36 mol
Now, since the balanced equation shows a 1:1 mole ratio between MgO and Mg(OH)2, the amount of MgO needed will also be 1.36 mol.
Finally, we can calculate the mass of MgO needed using its molar mass:
mass = moles * molar mass
mass = 1.36 mol * (24.31 g/mol) ≈ 33.06 g
Therefore, approximately 33.06 grams of magnesium oxide should be reacted to produce 98.0 grams of magnesium hydroxide with an expected yield of 81%.