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?
MgO(s)+H2O(l)->Mg(OH)2(aq)
Well, isn't magnesium hydroxide just being a little demanding here? Wanting a specific amount and expecting a certain yield? Talk about high expectations!
But fear not, my friend! I am here to help you navigate through these chemistry conundrums, even if it means applying some humorous logic. So let's dive into the world of magnesium hydroxide production!
To calculate the amount of magnesium oxide needed, we need to consider the expected yield of 81%. Essentially, the expected yield tells us that we will only be able to produce 81% of the total amount of magnesium hydroxide we expect.
Now, you want to produce 98.0 grams of magnesium hydroxide. If that's 100% of the desired amount, then only 81% will actually be produced. So we need to find the amount of magnesium oxide that will result in that 81% yield.
Let's do some math! If 81% of the target amount is 98.0 grams, then we can set up the following equation:
(81/100) x (Amount of Magnesium Oxide) = 98.0 grams
To find the amount of magnesium oxide, we can rearrange the equation:
Amount of Magnesium Oxide = (98.0 grams x 100) / 81
Now, let's input those numbers into our clown-calculator:
Amount of Magnesium Oxide = (98.0 x 100) / 81 ≈ You'll need approximately 120.99 grams of magnesium oxide!
So there you have it! To produce 98.0 grams of magnesium hydroxide with an expected yield of 81%, you'll need to react about 120.99 grams of magnesium oxide. Good luck with your chemistry adventures, my friend!
To calculate the amount of magnesium oxide required, we can use the molar mass and stoichiometry of the reaction.
1. Find the molar mass of magnesium hydroxide (Mg(OH)2):
Mg: 24.31 g/mol
O: 16.00 g/mol (2 atoms)
H: 1.01 g/mol (2 atoms)
Total molar mass: 24.31 + 2 × 16.00 + 2 × 1.01 = 58.33 g/mol
2. Determine the theoretical yield of magnesium hydroxide:
The expected yield is given as 81%, so we can multiply this by the amount we want (98.0 g) to find the theoretical yield:
Theoretical yield = 0.81 × 98.0 g = 79.38 g
3. Use the stoichiometry of the reaction to determine the moles of magnesium hydroxide:
According to the balanced equation, 1 mole of MgO reacts to produce 1 mole of Mg(OH)2.
Moles of Mg(OH)2 = mass / molar mass = 79.38 g / 58.33 g/mol ≈ 1.36 mol
4. Determine the moles of magnesium oxide:
Since the reaction is happening with excess water, all the moles of magnesium hydroxide produced will come from magnesium oxide.
Moles of MgO = 1.36 mol
5. Convert moles of MgO to grams:
Molar mass of MgO = 24.31 g/mol
Mass of MgO = moles × molar mass = 1.36 mol × 24.31 g/mol ≈ 33.02 g
Therefore, you should react approximately 33.02 g of magnesium oxide to produce 98.0 g of magnesium hydroxide with an expected yield of 81%.
To calculate the amount of magnesium oxide required to produce a given amount of magnesium hydroxide with a known yield, follow these steps:
Step 1: Write down the balanced chemical equation.
MgO(s) + H2O(l) -> Mg(OH)2(aq)
Step 2: Determine the molar mass of Mg(OH)2.
- Magnesium (Mg) has a molar mass of 24.31 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
- Hydrogen (H) has a molar mass of 1.01 g/mol.
Therefore, the molar mass of Mg(OH)2 is:
(24.31 g/mol) + 2(1.01 g/mol) + 2(16.00 g/mol) = 58.32 g/mol
Step 3: Calculate the molar mass of MgO.
- Magnesium (Mg) has a molar mass of 24.31 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
Therefore, the molar mass of MgO is:
24.31 g/mol + 16.00 g/mol = 40.31 g/mol
Step 4: Calculate the molar mass of water (H2O).
- Hydrogen (H) has a molar mass of 1.01 g/mol.
- Oxygen (O) has a molar mass of 16.00 g/mol.
Therefore, the molar mass of H2O is:
2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Step 5: Convert the given mass of magnesium hydroxide (98.0 g) to moles.
- Divide the given mass by the molar mass of Mg(OH)2:
98.0 g Mg(OH)2 / 58.32 g/mol = 1.68 mol Mg(OH)2
Step 6: Determine the theoretical yield of magnesium hydroxide.
Since the expected yield is 81%, only 81% of the calculated amount in Step 5 will actually be obtained:
1.68 mol Mg(OH)2 x 0.81 = 1.36 mol Mg(OH)2
Step 7: Use stoichiometry to find the amount of magnesium oxide required.
Refer to the balanced equation: 1 mol MgO reacts with 1 mol Mg(OH)2.
Therefore, the moles of MgO required will be the same as the moles of Mg(OH)2:
1.36 mol MgO
Step 8: Convert the moles of MgO to grams.
- Multiply the moles by the molar mass of MgO:
1.36 mol MgO x 40.31 g/mol = 54.79 g
Therefore, to produce 98.0 g of magnesium hydroxide with an expected yield of 81%, 54.79 grams of magnesium oxide should be reacted.
MgO(s)+H2O(l)->Mg(OH)2(aq)
You want 98.9 g Mg(OH)2. How many mols is that? mols = g/molar mass = approx 1.5 but you need to recalculate all of these estimates to get a better figure.
Convert mols Mg(OH)2 to mols Mg) using the coefficients in the balanced equation above. That is a 1:1 ratio; therefore mols MgO = approx 1.5.
Now convert mols MgO to grams MgO. That is mols MgO x molar mass MgO = approx 70g. That gives you the mass MgO needed in a 100% yield problem. It is is only 81% then you need about 70/0.81 = ? g MgO.
Post your work if you get stuck.