0
2 Mg (s) + O2 (g) → 2 MgO (s)
4.00mg x 1 mol Mg/24.305 x 2 mol MgO/2 mol Mg x 40.305g MgO/1 mol MgO=
=6.63g
4.00g- Given
24.305g- Atomic Mass of Magnesium
40.305g- Atomic Mass of Magnesium Oxide
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When 2.4 g of magnesium burns in excess oxygen 3.6 g of magnesium oxide is formed What is the percentage yield?
To calculate the percentage yield, we first need to determine the theoretical yield of magnesium oxide (MgO) from the given amount of magnesium. The molar mass of magnesium is approximately 24.3 g/mol, and that of magnesium oxide is about 40.3 g/mol. The balanced reaction shows that 2 moles of Mg produce 2 moles of MgO, meaning 2.4 g of Mg should theoretically yield about 3.6 g of MgO. Since the actual yield is also 3.6 g, the percentage yield is calculated as (actual yield/theoretical yield) × 100%, which results in a percentage yield of 100%.
What is the percent yield if the actual yield from this reaction is 306g?
To calculate the percent yield, you need the theoretical yield of the reaction. The percent yield is calculated using the formula: [ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 ] If you provide the theoretical yield, I can help you determine the percent yield.
What would be the products of 2Mg plus CO2?
The reaction between magnesium (Mg) and carbon dioxide (CO2) produces magnesium oxide (MgO) and carbon (C). The balanced chemical equation for this reaction is: 2Mg + CO2 → 2MgO + C. This indicates that two moles of magnesium react with one mole of carbon dioxide to yield two moles of magnesium oxide and one mole of carbon.
How do you calculate percentage yield?
# Determine the limiting reagent; # Calculate the expected yield if the reaction goes to 100% completion. # Divide the actual yield by the expected yield and multiply by 100. The result is percentage yield.
Is experimental yield the same as percent yield or actual yield?
Experimental yield and actual yield refer to the same thing, which is the amount of product obtained from a chemical reaction in a laboratory setting. Percent yield, on the other hand, is a measure of the efficiency of a reaction and is calculated by comparing the actual yield to the theoretical yield.
When 2.4 g of magnesium burns in excess oxygen 3.6 g of magnesium oxide is formed What is the percentage yield?
To calculate the percentage yield, we first need to determine the theoretical yield of magnesium oxide (MgO) from the given amount of magnesium. The molar mass of magnesium is approximately 24.3 g/mol, and that of magnesium oxide is about 40.3 g/mol. The balanced reaction shows that 2 moles of Mg produce 2 moles of MgO, meaning 2.4 g of Mg should theoretically yield about 3.6 g of MgO. Since the actual yield is also 3.6 g, the percentage yield is calculated as (actual yield/theoretical yield) × 100%, which results in a percentage yield of 100%.
In Part A we saw that the theoretical yield of aluminum oxide is 1.70 mol . Calculate the percent yield if the actual yield of aluminum oxide is 1.24 mol .?
Percent yield is calculated by dividing the actual yield by the theoretical yield, and then multiplying by 100. In this case, the percent yield would be: (1.24 mol / 1.70 mol) * 100 = 73%. This means that 73% of the theoretical yield was obtained in the experiment.
How can you achieve highest yield and best purity when burning magnesium to make magnesium oxide?
Use pure magnesium and pure oxygen.
Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product what would be the theoretical yield of the product?
2Mg + O2 --> 2MgO First, take the amount given (2.033g Mg) and convert it to moles. 1 mole of Mg weighs 24.312g. Then, convert to the desired element, which is MgO. Then convert the moles of MgO into grams. The complete problem looks like this: 2.033g Mg(1 mol Mg/24.312g Mg)(2 mol MgO/2 mol Mg)(40.311g MgO/1 mol MgO) = 3.371 g MgO 3.371 g MgO should be produced.
What is the percent yield for the reaction if 65.0 g of iron lll oxide produces 15.0g of iron?
Percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. The theoretical yield can be calculated by stoichiometry. In this case, the theoretical yield of iron would be 43.3g. Therefore, the percent yield would be (15.0g / 43.3g) x 100 = 34.6%.
What is the theoretical yield of aluminum oxide if 3.00 mol of aluminum metal is exposed to 2.55 mol of oxygen?
The balanced chemical equation for the reaction is 4Al + 3O2 -> 2Al2O3. Since the molar ratio between aluminum and oxygen is 4:3, aluminum is the limiting reagent. Therefore, the theoretical yield of aluminum oxide is based on the 3.00 mol of aluminum. Using stoichiometry, the theoretical yield of aluminum oxide would be 3.00 mol of Al * (2 mol Al2O3 / 4 mol Al) = 1.50 mol of Al2O3.
Copper dissolves in nitric acid producing copper nitrate After a reaction with sodium hydroxide and heating copper nitrate changed to copper oxide Calculate the percent yield?
To calculate the percent yield, you need to know the amount of copper oxide formed and compare it to the theoretical yield. The theoretical yield can be calculated based on the initial amount of copper, assuming complete conversion. Once you have both values, use the formula: Percent Yield = (Actual Yield / Theoretical Yield) x 100.
What is the difference between actual yield and theoretical yield?
Theoretical= calculated
What is the theoretical yield of aluminum oxide if 3.40 mol of aluminum metal is exposed to 2.85 mol of oxygen?
The balanced equation for the reaction is 4Al + 3O2 -> 2Al2O3. From the stoichiometry of the reaction, 3.40 mol of Al will react with 2.55 mol of O2 to produce 4.25 mol of Al2O3. Hence, the theoretical yield of aluminum oxide is 4.25 mol.
What is the theoretical yield of aluminum oxide if 3.40 of aluminum metal is exposed to 2.85 of oxygen?
To determine the theoretical yield of aluminum oxide, we first need to write and balance the chemical equation: 4 Al + 3 O2 -> 2 Al2O3. From the equation, we can see that 4 moles of Al reacts with 3 moles of O2 to produce 2 moles of Al2O3. Calculate the moles of Al and O2 provided, convert each to moles of Al2O3, and find the limiting reactant to determine the theoretical yield.
Is actual yield greater or less than theoretical yield?
The actual yield is less than the theoretical yield.
In the 1800's how did they extract potassium and magnesium?
Magnesium is extracted from its ores by one of two processes. In the first, the ore is converted to magnesium chloride (MgCl2), which is then electrolyzed. In the second process, the ore is converted to magnesium oxide (MgO), which is then treated with the alloy ferrosilicon. The ferrosilicon reacts with magnesium oxide to yield pure magnesium metal.It is the same for potassium, they both need to be electrolyised
