Barium sulfate forms as a precipitate in the following reaction:

Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNo3 (aq)
When 35.0g of Ba(nO3)s is reacted with excess Na2SO4, 29.8g of BaSO4 is recovered by the chemist.

a) Calculate the theoretical yield of BaSO4.
b) Calculate the percentage yield of BaSO4.

note: I'm not sure what I do because two of the grams -35.0g and 29.8g - are confusing me. I tried finding the molar mass. Now what?

The answer for
a) 31.3g
b) 95.2%

But I don't understand how to get it. Please explain, thank you.

This is my problem too

Please I need full working out in order to understand

To calculate the theoretical yield of BaSO4, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

Let's calculate the moles of Ba(NO3)2 and Na2SO4 using their molar masses:

Molar mass of Ba(NO3)2 = 261.34 g/mol
Molar mass of Na2SO4 = 142.04 g/mol

Moles of Ba(NO3)2 = 35.0 g / 261.34 g/mol = 0.134 mol
Moles of Na2SO4 = 29.8 g / 142.04 g/mol = 0.210 mol

From the balanced equation, we can see that the ratio of moles between Ba(NO3)2 and BaSO4 is 1:1. Therefore, the moles of BaSO4 that would be formed should also be 0.134 mol.

To calculate the theoretical yield of BaSO4 in grams, we can use its molar mass:

Molar mass of BaSO4 = 233.39 g/mol

Theoretical yield of BaSO4 = 0.134 mol * 233.39 g/mol = 31.3 g

Therefore, the theoretical yield of BaSO4 is 31.3 grams.

To calculate the percentage yield of BaSO4, we need to compare the actual yield (29.8 g) to the theoretical yield (31.3 g).

Percentage yield = (actual yield / theoretical yield) * 100
= (29.8 g / 31.3 g) * 100
= 95.2%

Therefore, the percentage yield of BaSO4 is 95.2%.

To solve this problem, we need to use stoichiometry and the concept of theoretical and percentage yield.

a) Calculating the theoretical yield of BaSO4:
1. Start by finding the molar mass of BaSO4.
- The molar mass of Ba(NO3)2 is 261.34 g/mol (atomic masses: Ba = 137.33 g/mol, N = 14.01 g/mol, O = 16.00 g/mol).
- The molar mass of BaSO4 is 233.39 g/mol (atomic masses: Ba = 137.33 g/mol, S = 32.07 g/mol, O = 16.00 g/mol).
2. Use the balanced equation to determine the stoichiometric ratio between Ba(NO3)2 and BaSO4. From the equation, we see that 1 mole of Ba(NO3)2 produces 1 mole of BaSO4.
3. Convert the given mass of Ba(NO3)2 to moles:
- Moles of Ba(NO3)2 = mass / molar mass
= 35.0 g / 261.34 g/mol
≈ 0.134 moles
4. Use the stoichiometric ratio to determine the moles of BaSO4:
- Moles of BaSO4 = moles of Ba(NO3)2
≈ 0.134 moles
5. Convert the moles of BaSO4 to mass by multiplying by the molar mass:
- Mass of BaSO4 = moles × molar mass
= 0.134 moles × 233.39 g/mol
≈ 31.3 g

Therefore, the theoretical yield of BaSO4 is approximately 31.3 grams.

b) Calculating the percentage yield of BaSO4:
1. Use the equation for percentage yield:
- Percentage yield = (Actual yield / Theoretical yield) × 100%
2. Given actual yield = 29.8 g and theoretical yield ≈ 31.3 g, substitute these values into the equation:
- Percentage yield = (29.8 g / 31.3 g) × 100%
= 95.2%

Therefore, the percentage yield of BaSO4 is 95.2%.

In conclusion:
a) The theoretical yield of BaSO4 is approximately 31.3 grams.
b) The percentage yield of BaSO4 is 95.2%.