A- 50.0 g of H2SO4 react with 75.0 g of NaOH. Identify the limiting and excess reactants. How many grams of Na2SO4 will be formed?

B- A chemist burns 160.0 g of aluminum in 200.0 g of oxygen. She produces 275.0 g of aluminum oxide. Which is the limiting reactant? Which is the excess reactant? How many grams of aluminum oxide should theoretically be produced? What is her percent yield?

1. How many moles are in 125g of H2SO4 ?
2. How many grams are in 1.5 x 1025 molecules of CO2 ?
3. Determine the percentage composition of KMnO4 .
4. What is the empirical formula of a compound that was found to consist of 22.1% aluminum, 25.4% phosphorous, 52.5% oxygen?
5. A compound is found to contain 18.7% lithium, 16.3% carbon and 65.0% oxygen. If the molar mass of the compound is 73.8 g/mol, what is the compound’s molecular formula?

6. How many grams of hydrogen are produced when 25.0 g of zinc reacts with excess Hydrochloric acid?
7. Silver nitrate and Sodium phosphate react in equal amounts of 200.0 g each. How many grams of Silver phosphate are produced?
8. If the reaction above is performed and yields only 155 g of Silver phosphate, what was the percent yield?

A- 50.0 g of H2SO4 react with 75.0 g of NaOH. Identify the limiting and excess reactants. How many grams of Na2SO4 will be formed?

I will work this ONE limiting reagent (LR) problem for you. That should get you started on the others. I work these the long way. Here are the steps.
1. Write and balance the equation.
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
2. Convert what you have into moles.
a. mols H2SO4 = grams/molar mass 50/98 = 0.51
b. mols NaOH = 75/40 = 1.875
3. Convert EACH of 2a and 2b into any product. You do this by using the coefficients in the balanced equation. I'll choose Na2SO4.
3a (from 2a.) 0.51 mols H2SO4 x (1 mol Na2SO4/1 mol H2SO4) = 0.51 mols Na2SO4 produced IF we had 50 g H2SO4 and all the NaOH we needed.
3b (from 2b). 1.875 mol NaOH x (1 mol Na2SO4/2 mol NaOH) = 0.938 mols Na2SO4 IF we had 75 g NaOH and all the H2SO4 we needed.
4. In LR problems the SMALL number always wins because you can't produce more than the smallest amount; therefore, 0.51 moles of H2SO4 is the LR, NaOH is the excess reagent.
5. grams Na2SO4 produced = mols Na2SO4 x molar mass Na2SO4 = 0.51 x 142 = ? This is called the theoretical yield (TY)
If you want percent yield the problem will tell you how much was produced, which I will call AY (for actual yield). Then
% yield = 100 x (AY/TY) = ?
Some problems will as for how much of the excess reagent remains. Just run a separate stoichiometry problem using the amount of H2SO4(in this case) to see how much NaOH is consumed and subtract from the initial amount to find the amount remaining.
Post your work if you get stuck on this or any of the others.

A- Well, first we need to find the molar masses of H2SO4 and NaOH. H2SO4 has a molar mass of approximately 98.09 g/mol, and NaOH has a molar mass of approximately 40.00 g/mol.

To determine the limiting reactant, we need to compare the number of moles of each reactant. For H2SO4, we have 50.0 g, which is approximately 0.51 moles. For NaOH, we have 75.0 g, which is approximately 1.88 moles.

Now, let's calculate the moles of NaOH needed to react with 0.51 moles of H2SO4. The balanced chemical equation shows us that the ratio is 1:1, so we need 0.51 moles of NaOH. Since we have 1.88 moles, NaOH is in excess.

To find the grams of Na2SO4 formed, we need to use the stoichiometry of the balanced equation. The balanced equation is:

H2SO4 + 2NaOH -> Na2SO4 + 2H2O.

From the equation, we see that 1 mole of H2SO4 produces 1 mole of Na2SO4. Thus, 0.51 moles of H2SO4 will produce 0.51 moles of Na2SO4.

To convert moles to grams, we multiply the moles of Na2SO4 by its molar mass. The molar mass of Na2SO4 is approximately 142.04 g/mol.

0.51 moles of Na2SO4 x 142.04 g/mol = 72.56 grams of Na2SO4.

So, 72.56 grams of Na2SO4 will be formed.

B- Oh, burning aluminum and oxygen, that's a hot combination! First, let's check the balanced equation for this reaction:

4Al + 3O2 -> 2Al2O3.

Now, let's calculate the number of moles for each reactant.

Aluminum has a molar mass of approximately 26.98 g/mol. So, for 160.0 g of aluminum, we have approximately 5.93 moles.

Oxygen has a molar mass of approximately 32.00 g/mol. So, for 200.0 g of oxygen, we have approximately 6.25 moles.

Since the balanced equation shows us that the ratio of aluminum to oxygen is 4:3, we can see that 5.93 moles of aluminum would require 4*(5.93/4) = 5.93 moles of oxygen. Since we only have 6.25 moles of oxygen, this means oxygen is in excess.

To determine the limiting reactant, we compare the moles of each reactant to the stoichiometry of the balanced equation. Since aluminum is the limiting reactant, we will calculate the grams of Al2O3 formed using the molar mass of aluminum oxide, which is approximately 101.96 g/mol.

Based on the balanced equation, we know that 4 moles of aluminum produces 2 moles of Al2O3. Therefore, 5.93 moles of aluminum would produce (5.93/4)*(2) = 2.96 moles of Al2O3.

To convert moles to grams, we multiply the moles of Al2O3 by its molar mass:

2.96 moles of Al2O3 x 101.96 g/mol = 301.81 grams of Al2O3.

So, theoretically, 301.81 grams of aluminum oxide should be produced.

As for the percent yield, we need to know the actual yield of aluminum oxide to calculate it. Since the question tells us that 275.0 grams of aluminum oxide were produced, we can now calculate the percent yield.

The percent yield is given by the formula:

Percent Yield = (Actual Yield / Theoretical Yield) * 100.

Plugging in the values:

Percent Yield = (275.0 g / 301.81 g) * 100 = 91.11%.

So, her percent yield is 91.11%.

Now let's move on to the next set of questions!

1. Ah, the moles game. We convert grams of H2SO4 to moles by using its molar mass, which is approximately 98.09 g/mol.

125 g of H2SO4 x (1 mol / 98.09 g) = 1.27 moles of H2SO4.

So, there are approximately 1.27 moles of H2SO4 in 125 grams.

2. To calculate the number of grams in a given number of molecules, we need to know the molar mass of CO2, which is approximately 44.01 g/mol.

1.5 x 10^25 molecules of CO2 x (44.01 g/mol / 6.022 x 10^23 molecules/mol) = 1104.89 grams of CO2.

Therefore, there are approximately 1104.89 grams in 1.5 x 10^25 molecules of CO2.

3. Percentage composition...sounds serious, but it's just like figuring out how a pizza is divided. The molar mass of KMnO4 is approximately 158.04 g/mol.

To find the mass of each element in 1 mole of KMnO4, we multiply the molar mass by the number of atoms of each element per molecule:

Mass of K: 1 * (39.10 g/mol) = 39.10 g.
Mass of Mn: 1 * (54.94 g/mol) = 54.94 g.
Mass of O: 4 * (16.00 g/mol) = 64.00 g.

Now, to find the percentage composition, we divide the mass of each element by the total molar mass of KMnO4 and multiply by 100:

Percentage of K: (39.10 g / 158.04 g) * 100 = 24.74%.
Percentage of Mn: (54.94 g / 158.04 g) * 100 = 34.80%.
Percentage of O: (64.00 g / 158.04 g) * 100 = 40.46%.

So, the percentage composition of KMnO4 is approximately 24.74% K, 34.80% Mn, and 40.46% O.

4. Empirical formula time! We have percentages, so we can assume a 100g sample of this compound to make it easier. Then we have 22.1g of aluminum, 25.4g of phosphorus, and 52.5g of oxygen.

To find the moles of each element, we divide the mass of each element by its molar mass:

Moles of Al: 22.1 g / 26.98 g/mol = 0.819 moles.
Moles of P: 25.4 g / 30.97 g/mol = 0.821 moles.
Moles of O: 52.5 g / 16.00 g/mol = 3.281 moles.

Since we want the simplest whole number ratio, we divide each number of moles by the smallest number of moles (0.819) and round to the nearest whole number:

Moles of Al: 0.819 moles / 0.819 moles = 1.
Moles of P: 0.821 moles / 0.819 moles = 1.
Moles of O: 3.281 moles / 0.819 moles = 4.

So, the empirical formula of the compound is AlPO4.

5. Molecules, percentages, and molar mass, oh my! First, we need to find the empirical formula based on the percentage composition.

Since we're given percentages, we'll assume a 100g sample of the compound. This means we have 18.7g of lithium, 16.3g of carbon, and 65.0g of oxygen.

To find the moles of each element, we divide the mass of each element by its molar mass:

Moles of Li: 18.7 g / 6.94 g/mol (lithium's molar mass) = 2.69 moles.
Moles of C: 16.3 g / 12.01 g/mol (carbon's molar mass) = 1.36 moles.
Moles of O: 65.0 g / 16.00 g/mol (oxygen's molar mass) = 4.06 moles.

Next, we divide each number of moles by the smallest number of moles (1.36) and round to the nearest whole number:

Moles of Li: 2.69 moles / 1.36 moles = 1.98 (approximately 2).
Moles of C: 1.36 moles / 1.36 moles = 1.
Moles of O: 4.06 moles / 1.36 moles = 2.99 (approximately 3).

So, the empirical formula of the compound is Li2CO3.

Now, let's move on to the last set of questions!

6. Well, this is a classic reaction between zinc and hydrochloric acid. To answer the question, we need to know the balanced equation:

Zn + 2HCl -> ZnCl2 + H2.

According to the equation, 1 mole of zinc produces 1 mole of hydrogen gas. To find the moles of hydrogen produced when 25.0 g of zinc reacts, we need to convert the mass of zinc to moles. The molar mass of zinc is approximately 65.38 g/mol:

25.0 g of zinc x (1 mol / 65.38 g) = 0.38 moles of zinc.

Since the ratio of zinc to hydrogen is 1:1, 0.38 moles of zinc will produce 0.38 moles of hydrogen.

To find the grams of hydrogen produced, we multiply the moles of hydrogen by its molar mass, which is approximately 2.02 g/mol:

0.38 moles of hydrogen x 2.02 g/mol = 0.77 grams of hydrogen.

Therefore, approximately 0.77 grams of hydrogen will be produced when 25.0 g of zinc reacts with excess hydrochloric acid.

7. Ah, the classic silver nitrate and sodium phosphate reaction. Let's check the balanced equation:

3AgNO3 + Na3PO4 -> Ag3PO4 + 3NaNO3.

According to the equation, 3 moles of silver phosphate are formed for every 3 moles of sodium phosphate used. Therefore, the number of grams of silver phosphate formed is equal to the number of grams of sodium phosphate used:

200.0 g of sodium phosphate will form 200.0 g of silver phosphate.

So, 200.0 grams of silver phosphate will be produced when 200.0 grams of sodium phosphate are used.

8. Percent yield strikes again! The theoretical yield is the amount of silver phosphate that should have been produced based on stoichiometry, which we determined to be 200.0 grams.

We know that the actual yield is only 155.0 grams of silver phosphate, as given in the question.

To find the percent yield, we use the formula:

Percent Yield = (Actual Yield / Theoretical Yield) * 100.

Plugging in the values:

Percent Yield = (155.0 g / 200.0 g) * 100 = 77.5%.

So, the percent yield is 77.5%.

A- To identify the limiting and excess reactants, you need to compare the amount of each reactant to the stoichiometry of the balanced chemical equation.

1. Convert the given masses of H2SO4 and NaOH to moles using their molar masses (H2SO4: 98.09 g/mol, NaOH: 40.00 g/mol).
- Moles of H2SO4 = 50.0 g / 98.09 g/mol = 0.510 mol
- Moles of NaOH = 75.0 g / 40.00 g/mol = 1.875 mol

2. Use the stoichiometry from the balanced chemical equation to determine the mole ratio between H2SO4 and NaOH. The balanced equation is:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O

The ratio of moles of H2SO4 to NaOH is 1:2. This means that 1 mole of H2SO4 reacts with 2 moles of NaOH.

3. Divide the number of moles calculated in step 1 by the stoichiometric ratio to determine the number of moles of NaOH required for complete reaction with the given amount of H2SO4.
- Moles of NaOH required = 0.510 mol H2SO4 * (2 mol NaOH / 1 mol H2SO4) = 1.02 mol

4. Compare the moles of NaOH available (1.875 mol) to the moles of NaOH required (1.02 mol). The reactant with fewer moles available is the limiting reactant, and the other reactant is in excess.
- Since 1.875 mol > 1.02 mol, NaOH is in excess, and H2SO4 is the limiting reactant.

5. To calculate the grams of Na2SO4 formed, you need to use the stoichiometry from the balanced equation.
- From the balanced equation, 1 mole of H2SO4 reacts with 1 mole of Na2SO4.
- The number of moles of Na2SO4 formed is equal to the number of moles of H2SO4 used since they have a 1:1 stoichiometric ratio.
- Moles of Na2SO4 formed = 0.510 mol

6. Convert the moles of Na2SO4 to grams using its molar mass (Na2SO4: 142.04 g/mol).
- Grams of Na2SO4 formed = 0.510 mol * 142.04 g/mol = 72.572 g

Therefore, the limiting reactant is H2SO4, the excess reactant is NaOH, and 72.572 grams of Na2SO4 will be formed.

B- To determine the limiting and excess reactants and calculate the theoretical yield and percent yield, follow these steps:

1. Convert the given masses of aluminum and oxygen to moles using their molar masses (Al: 26.98 g/mol, O2: 32.00 g/mol).
- Moles of Al = 160.0 g / 26.98 g/mol = 5.93 mol
- Moles of O2 = 200.0 g / 32.00 g/mol = 6.25 mol

2. Use the stoichiometry from the balanced chemical equation to determine the mole ratio between Al and O2. The balanced equation is:
4Al + 3O2 -> 2Al2O3

The ratio of moles of Al to O2 is 4:3. This means that 4 moles of Al react with 3 moles of O2.

3. Divide the number of moles calculated in step 1 by the stoichiometric ratio to determine the number of moles of O2 required for complete reaction with the given amount of Al.
- Moles of O2 required = 5.93 mol Al * (3 mol O2 / 4 mol Al) = 4.4475 mol

4. Compare the moles of O2 available (6.25 mol) to the moles of O2 required (4.4475 mol). The reactant with fewer moles available is the limiting reactant, and the other reactant is in excess.
- Since 6.25 mol > 4.4475 mol, O2 is in excess, and Al is the limiting reactant.

5. To calculate the theoretical yield of Al2O3, use the stoichiometry from the balanced equation.
- From the balanced equation, 4 moles of Al react to form 2 moles of Al2O3.
- The number of moles of Al2O3 formed is half the number of moles of Al used since they have a 4:2 stoichiometric ratio.
- Moles of Al2O3 formed = 5.93 mol Al / 2 = 2.965 mol

6. Convert the moles of Al2O3 to grams using its molar mass (Al2O3: 101.96 g/mol).
- Grams of Al2O3 formed = 2.965 mol * 101.96 g/mol = 302.03 g

Therefore, the limiting reactant is Al, the excess reactant is O2, the theoretical yield of Al2O3 is 302.03 grams, and to calculate the percent yield, divide the actual yield by the theoretical yield and multiply by 100.