magnesium powder reacts with steam to form magnesium hydroxide and hydrogen gas. What is the pecentage yield of 10.1g Mg reacts with an excess of water abnd 21.0g Mg 9g (OH)2 is recovered?

If 24g Mg is used the percentage yield is 95% how many gramns of magnesium hydroxide should be recovered?

You REALLY should read questions you post. The typos make this almost useless; however, I will assume you meant to write, ".....and 21.0 g Mg(OH)2 is recovered?"

Mg + 2HOH ==> Mg(OH)2 + H2
1. Convert 10.1 g Mg to moles. Moles = grams/atomic mass.
2. Using the coefficients in the balanced equation, convert moles Mg to moles Mg(OH)2.
3. Now convert moles Mg(OH)2 to grams. grams = moles x molar mass.The value, in grams, which you obtain here is the THEORETICAL YIELD.
4. Now that you have the theoretical yield, you can calculate the percent yield.
%yield = (grams obtained/theoretical yield)*100 =??

For the second part, start with 24 g Mg and determine the theoretical yield. Then multiply by 0.95 to obtain the actual yield if the process is 95% efficient.

What is the percentage yield of 10.1g mg reacts with an excess of water and 21.0 g Mg(OH)2 is removed ?

To find the percentage yield, you need to compare the actual yield (the amount of product obtained) with the theoretical yield (the maximum amount of product that can be obtained).

1. Start by finding the molar mass of magnesium (Mg) and magnesium hydroxide (Mg(OH)2):
- Mg: 24.31 g/mol
- Mg(OH)2: 58.33 g/mol

2. Calculate the moles of magnesium (Mg):
Given mass of Mg = 10.1 g
Moles of Mg = mass / molar mass = 10.1 g / 24.31 g/mol = 0.4158 mol

3. Determine the mole ratio between Mg and Mg(OH)2 using the balanced chemical equation:
Mg(s) + 2H2O(g) -> Mg(OH)2(aq) + H2(g)
From the equation, it can be seen that 1 mole of Mg reacts with 1 mole of Mg(OH)2.
Therefore, the moles of Mg(OH)2 should be equal to the moles of Mg, which is 0.4158 mol.

4. Calculate the theoretical yield (the maximum amount of Mg(OH)2 that can be produced):
Mass of Mg(OH)2 = moles x molar mass = 0.4158 mol x 58.33 g/mol = 24.245 g

5. Calculate the percentage yield:
Given mass of Mg(OH)2 = 21.0 g
Percentage yield = (actual yield / theoretical yield) x 100%
Percentage yield = (21.0 g / 24.245 g) x 100% = 86.5%

Therefore, the percentage yield of the reaction when 10.1 g of Mg reacts is 86.5%.

To determine the grams of Mg(OH)2 that should be recovered when 24 g of Mg is used (with a 95% yield):

1. Calculate the theoretical yield:
Mass of Mg = 24 g
Moles of Mg = mass / molar mass = 24 g / 24.31 g/mol = 0.987 mol (approximately)
Moles of Mg(OH)2 = moles of Mg = 0.987 mol
Mass of Mg(OH)2 = moles x molar mass = 0.987 mol x 58.33 g/mol = 57.46 g (approximately)

Therefore, when 24 g of Mg is used with a 95% yield, approximately 57.46 g of Mg(OH)2 should be recovered.

To calculate the percentage yield, you first need to determine the theoretical yield, which is the maximum amount of product that can be obtained based on the balanced chemical equation.

1. Start by balancing the chemical equation for the reaction between magnesium and steam:

Mg + H2O -> Mg(OH)2 + H2

2. Determine the molar masses of the substances involved:

Mg: 24.31 g/mol
H2O: 18.02 g/mol
Mg(OH)2: 58.33 g/mol
H2: 2.02 g/mol

3. Convert the given mass of magnesium (10.1g) to moles by dividing by its molar mass:

Moles of Mg = 10.1 g / 24.31 g/mol

4. Based on the balanced equation, you can establish a stoichiometric ratio between magnesium and magnesium hydroxide. From the equation, you can see that 1 mole of Mg reacts to form 1 mole of Mg(OH)2. Therefore, the number of moles of Mg(OH)2 can be determined as:

Moles of Mg(OH)2 = Moles of Mg

5. Calculate the mass of magnesium hydroxide that should be obtained using its molar mass:

Mass of Mg(OH)2 = Moles of Mg(OH)2 * Molar mass of Mg(OH)2

6. Finally, calculate the percentage yield using the actual yield (the given mass of Mg(OH)2, 21.0g) and the theoretical yield (calculated in step 5):

Percentage yield = (Actual yield / Theoretical yield) * 100%

Now let's calculate the percentage yield for the given values:

1. For the first reaction with 10.1g of Mg and 21.0g of Mg(OH)2:

Theoretical yield of Mg(OH)2 = Moles of Mg(OH)2 * Molar mass of Mg(OH)2
Theoretical yield of Mg(OH)2 = (10.1 g / 24.31 g/mol) * 58.33 g/mol

Percentage yield = (21.0 g / (10.1 g / 24.31 g/mol * 58.33 g/mol)) * 100%

2. For the second reaction with 24g of Mg and the percentage yield of 95%:

Theoretical yield of Mg(OH)2 = Moles of Mg(OH)2 * Molar mass of Mg(OH)2
Theoretical yield of Mg(OH)2 = (24 g / 24.31 g/mol) * 58.33 g/mol

Actual yield of Mg(OH)2 = Theoretical yield of Mg(OH)2 * Percentage yield

You can use these calculations to find the respective mass of magnesium hydroxide for each scenario.