A limiting reactants problem.
6.32g of sodium sulfate is reacted with 12.03g of barium nitrate. How many grams of the precipitate would you expect to collect.
The answer my teacher gave me is 16.0g BaSO4, but I cannot get it.
The equation I formed is:
Na2SO4 + Ba(NO3)2 --> 2NaNO3 + BaSO4
Thanks in advance!
I think 16.0 g BaSO4 is wrong.
I work these the long way.
First we take mols Na2SO4 and all the Ba(NO3)2 needed and solve for mols BaSO4 formed.
Next we take mols Ba(NO)3 and all the Na2SO4 needed and solve for mols BaSO4 formed.
The smaller number of mols BaSO4 formed wins.
mols Na2SO4 = 6.32/142 = 0.0445
That gives you 0.0445 mols BaSO4 and convert to g BaSO4 = 0.0445*233.39 = 10.4g
mols Ba(NO3)2 = 12.03/261.337 = 0.046
0.046*233.39 = 10.7 g
10.4g BaSO4 is what you will obtain. How that compare with your value?
I got 10.4g as well. My teacher must have put down the incorrect answer... thank you for your help!
To solve this limiting reactants problem, we need to determine which reactant is the limiting reactant and calculate the amount of the precipitate formed.
Step 1: Calculate the moles of each reactant.
Moles of Na2SO4 = mass / molar mass = 6.32g / (22.99 x 2 + 32.07 + 16.00 x 4) = 0.042 mol
Moles of Ba(NO3)2 = mass / molar mass = 12.03g / (137.33 + 3 x 14.01 + 2 x 16.00) = 0.040 mol
Step 2: Determine the stoichiometric ratio.
From the balanced equation, the ratio between Na2SO4 and BaSO4 is 1:1. This means that for every 1 mole of Na2SO4, 1 mole of BaSO4 is produced.
Step 3: Identify the limiting reactant.
To determine the limiting reactant, compare the moles of each reactant to their stoichiometric ratio. The reactant that produces fewer moles of the desired product is the limiting reactant.
In this case, both reactants have a 1:1 stoichiometric ratio with BaSO4. However, Na2SO4 has a higher mole quantity (0.042 mol) compared to Ba(NO3)2 (0.040 mol). Therefore, Ba(NO3)2 is the limiting reactant.
Step 4: Calculate the mass of BaSO4 formed.
Since we know that barium nitrate (Ba(NO3)2) is the limiting reactant, we can use its moles to determine the amount of BaSO4 formed.
Moles of BaSO4 formed = Moles of limiting reactant (Ba(NO3)2) = 0.040 mol
Now, we can calculate the mass of BaSO4 formed using the moles and molar mass of BaSO4:
Mass of BaSO4 formed = Moles of BaSO4 x molar mass of BaSO4
= 0.040 mol x (137.33 + 32.07 + 16.00 x 4)
= 0.040 mol x 233.39 g/mol
= 9.3356 g ≈ 9.34 g
Therefore, the expected mass of the precipitate, BaSO4, is approximately 9.34 grams, not 16.0 grams as your teacher mentioned.
To solve this limiting reactants problem, we need to determine the limiting reactant first. The limiting reactant is the reactant that will be completely consumed in the reaction and limits the amount of product formed.
To find the limiting reactant, we need to calculate the number of moles of each reactant. We can do this by using the molar mass of each compound:
Molar mass of Na2SO4:
2(22.99 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 142.04 g/mol
Molar mass of Ba(NO3)2:
137.33 g/mol + 2(14.01 g/mol) + 6(16.00 g/mol) = 261.34 g/mol
Now, let's calculate the number of moles for each reactant:
Number of moles of Na2SO4 = 6.32 g / 142.04 g/mol = 0.0445 mol
Number of moles of Ba(NO3)2 = 12.03 g / 261.34 g/mol = 0.0460 mol
According to the balanced equation, the stoichiometric ratio between Na2SO4 and BaSO4 is 1:1. This means that for every mole of Na2SO4 reacted, 1 mole of BaSO4 is produced.
Since the molar ratio of Na2SO4 and BaSO4 is 1:1, the limiting reactant will be the one with fewer moles. In this case, Na2SO4 has fewer moles (0.0445 mol), so it is the limiting reactant.
Now that we have identified the limiting reactant, we can use its mole ratio to calculate the moles and then the grams of BaSO4 produced.
From the balanced equation:
1 mol Na2SO4 reacts with 1 mol BaSO4
So, the number of moles of BaSO4 formed will be equal to the number of moles of Na2SO4 reacted, which is 0.0445 mol.
Now, let's calculate the mass of BaSO4 produced:
Mass of BaSO4 = number of moles of BaSO4 x molar mass of BaSO4
Mass of BaSO4 = 0.0445 mol x (137.33 g/mol + 32.07 g/mol + 4(16.00 g/mol))
Mass of BaSO4 = 0.0445 mol x 233.40 g/mol
Mass of BaSO4 = 10.37 g
Therefore, the expected mass of the precipitate (BaSO4) that you would collect is 10.37 grams.
It seems that the answer your teacher provided (16.0g BaSO4) may be incorrect. Double-check your calculations and make sure you have the correct molar masses and stoichiometric ratios.