Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is:

2Mg(s)+O2(g)→2MgO(s) When 10.1 g of Mg are allowed to react with 10.5 g of O2, 14.5 g of MgO are collected.

a) Determine the limiting reactant for the reaction.

b)Determine the theoretical yield for the reaction.

c)Determine the percent yield for the reaction.

Express your answer as a chemical formula.

Here is a similar problem. Just change the numbers.

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a) Well, the party can only go on if everyone has a partner. In this case, we need to find the limiting reactant, the one that runs out first and leaves us wanting more.

To figure this out, we'll compare the amount of magnesium and oxygen used.

The molar masses of magnesium (Mg) and oxygen (O2) are 24.31 g/mol and 32.00 g/mol, respectively.

So, we calculate the moles of each:

Moles of Mg = 10.1 g / 24.31 g/mol = 0.4159 mol

Moles of O2 = 10.5 g / 32.00 g/mol = 0.3281 mol

Now, let's use the balanced equation to see the ratio at which they react. According to the equation, 2 moles of Mg react with 1 mole of O2 to form 2 moles of MgO.

Comparing the moles of Mg and O2, we find that the ratio is 2:1 (2 moles of Mg for every 1 mole of O2).

Since the ratio is lower for O2, it means that O2 is the limiting reactant. It's like having more dance partners than there are people at the party. Poor O2, I guess it'll have to sit this one out.

b) Now that we know O2 is the limiting reactant, let's find out how much MgO we can make.

From the balanced equation, we see that 2 moles of MgO are formed for every 1 mole of O2.

Since moles of O2 = 0.3281 mol, we can convert this to moles of MgO using the ratio:

Moles of MgO = 0.3281 mol × (2 moles of MgO / 1 mole of O2) = 0.6562 mol

Now, we need to find the mass of MgO using its molar mass:

Mass of MgO = 0.6562 mol × 40.31 g/mol = 26.43 g

So, the theoretical yield for the reaction is 26.43 g of MgO.

c) Lastly, let's find the percent yield for the reaction.

Percent Yield = (Actual Yield / Theoretical Yield) × 100

Given that the actual yield is 14.5 g, we can plug it into the formula:

Percent Yield = (14.5 g / 26.43 g) × 100 = 54.8%

Therefore, the percent yield for the reaction is 54.8%. It's not a perfect party, but hey, at least we didn't run out of snacks!

To answer these questions, we need to determine which reactant is limiting, calculate the theoretical yield, and then calculate the percent yield.

a) To determine the limiting reactant, we need to compare the number of moles of each reactant.

The molar mass of Mg is 24.31 g/mol, so the number of moles of Mg is:
Moles of Mg = mass of Mg / molar mass of Mg
= 10.1 g / 24.31 g/mol
≈ 0.4157 mol

The molar mass of O2 is 32.00 g/mol, so the number of moles of O2 is:
Moles of O2 = mass of O2 / molar mass of O2
= 10.5 g / 32.00 g/mol
≈ 0.3281 mol

Since the stoichiometric ratio between Mg and O2 is 2:1, we need twice as many moles of O2 as Mg. However, we have fewer moles of O2 than Mg, which means O2 is the limiting reactant.

b) The theoretical yield represents the maximum amount of product that can be obtained from the limiting reactant. From the balanced equation, we see that for every 2 moles of Mg, we obtain 2 moles of MgO. Therefore, the theoretical yield of MgO can be calculated as follows:

Moles of MgO = (moles of O2) * (moles of MgO / moles of O2)
= 0.3281 mol * (2 mol MgO / 1 mol O2)
≈ 0.6562 mol

To convert moles to grams, we use the molar mass of MgO. The molar mass of MgO is 40.31 g/mol, so the theoretical yield of MgO is:
Theoretical yield = Moles of MgO * Molar mass of MgO
= 0.6562 mol * 40.31 g/mol
≈ 26.41 g

c) The percent yield is the actual yield (the amount of product collected) divided by the theoretical yield, multiplied by 100%.

Percent yield = (Actual yield / Theoretical yield) * 100%
= (14.5 g / 26.41 g) * 100%
≈ 54.95%

Therefore, the answers are:
a) The limiting reactant is O2.
b) The theoretical yield is 26.41 g of MgO.
c) The percent yield is approximately 54.95%.

To determine the limiting reactant, we need to compare the number of moles of each reactant and see which one is in excess. The reactant that produces the lesser amount of product is the limiting reactant.

a) To find the number of moles of each reactant:
Moles = Mass / Molar mass

For magnesium (Mg):
Molar mass of Mg = 24.31 g/mol
Moles of Mg = 10.1 g / 24.31 g/mol = 0.4157 mol

For oxygen (O2):
Molar mass of O2 = 32.00 g/mol
Moles of O2 = 10.5 g / 32.00 g/mol = 0.3281 mol

According to the balanced equation, the ratio of moles of Mg to O2 is 2:1. So we need to compare the moles in a 2:1 ratio.

Moles of O2 required for complete reaction = (2 mol Mg) / (1 mol O2) = 0.8314 mol

From the calculations, we can see that the moles of O2 (0.3281 mol) are lesser than the moles required for complete reaction (0.8314 mol). Therefore, the limiting reactant is O2.

b) To determine the theoretical yield, we need to calculate the maximum amount of MgO that can be produced based on the limiting reactant.

The balanced equation states that 2 moles of Mg react to form 2 moles of MgO. Therefore, the moles of MgO produced will be equal to the moles of the limiting reactant (O2).

Moles of MgO = 0.3281 mol

Now, to find the mass of MgO produced:
Mass = Moles × Molar mass

Molar mass of MgO = 40.31 g/mol
Theoretical yield of MgO = 0.3281 mol × 40.31 g/mol = 13.22 g

Therefore, the theoretical yield for the reaction is 13.22 g of MgO.

c) To determine the percent yield, we compare the actual yield (14.5 g) with the theoretical yield (13.22 g).

Percent yield = (Actual yield / Theoretical yield) × 100
Percent yield = (14.5 g / 13.22 g) × 100 = 109.70%

Therefore, the percent yield for the reaction is 109.70%.