In the lab, 20.0 g of sodium hydroxide is added to a beaker giving it a total mass of 128.4 g. Sulfuric acid is added until all of the sodium hydroxide reacts. After heating, the total mass of the beaker and the dry compound is 142.3 g. Caculate the percent yeild.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
Convert 20.0 g NaOH to mols NaOH.
Convert mols NaOH to mols Na2SO4 using the coefficients in the balanced equation above.
Convert mols Na2SO4 to grams Na2SO4 using mols x molar mass = grams. This value is the theoretical yield (in grams)
%yield = (A/T)*100
where A = actual mass Na2SO4 collected.
T = theoretical yield.
Post your work if you get stuck and need further assistance.
1. Convert 20.0 g NaOH to mols NaOH:
To do this, we need to find the molar mass of NaOH. The molar mass of sodium (Na) is 22.99 g/mol, the molar mass of oxygen (O) is 16.00 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol.
Molar mass of NaOH = (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol) = 39.00 g/mol
Now, we can calculate the number of moles (mols) of NaOH:
Mols NaOH = Mass of NaOH / Molar mass of NaOH
Mols NaOH = 20.0 g / 39.00 g/mol
Mols NaOH ≈ 0.513 mols NaOH
2. Convert mols NaOH to mols Na2SO4 using the coefficients in the balanced equation above:
From the balanced equation, we can see that the ratio of NaOH to Na2SO4 is 2:1. So, the molar ratio is 2 mols NaOH: 1 mol Na2SO4.
Therefore, the number of moles (mols) of Na2SO4 formed would be half the mols of NaOH:
Mols Na2SO4 = 0.513 mols NaOH / 2
Mols Na2SO4 ≈ 0.2575 mols Na2SO4
3. Convert mols Na2SO4 to grams Na2SO4 using mols x molar mass = grams:
The molar mass of Na2SO4 is the sum of the molar masses of its constituent elements: 2 mols sodium (Na), 1 mol sulfur (S), and 4 mols oxygen (O).
Molar mass of Na2SO4 = (2 x 22.99 g/mol) + (32.07 g/mol) + (4 x 16.00 g/mol) = 142.04 g/mol
Now, we can calculate the theoretical yield (in grams) of Na2SO4:
Theoretical yield = Mols Na2SO4 x Molar mass of Na2SO4
Theoretical yield = 0.2575 mols Na2SO4 x 142.04 g/mol
Theoretical yield ≈ 36.59 g Na2SO4
4. Calculate the percent yield:
%yield = (A / T) x 100
Where:
A = actual mass of Na2SO4 collected = (total mass of beaker and dry compound) - (mass of beaker)
T = theoretical yield of Na2SO4
Given that the total mass of the beaker and the dry compound is 142.3 g and the mass of the beaker is 128.4 g, we can find the actual mass of Na2SO4:
A = 142.3 g - 128.4 g
A = 13.9 g
Now, we can calculate the percent yield:
%yield = (13.9 g / 36.59 g) x 100
%yield ≈ 38.0%
Therefore, the percent yield of the reaction is approximately 38.0%.
To calculate the percent yield, we need to find the theoretical yield and the actual yield.
Step 1: Convert 20.0 g of NaOH to moles of NaOH.
The molar mass of NaOH is:
23.0 g/mol (Na) + 16.0 g/mol (O) + 1.0 g/mol (H) = 40.0 g/mol
So, the number of moles of NaOH is:
20.0 g / 40.0 g/mol = 0.5 moles
Step 2: Use the balanced equation to determine the moles of Na2SO4 that should be produced.
The balanced equation shows that 2 moles of NaOH react to form 1 mole of Na2SO4.
So, the number of moles of Na2SO4 is:
0.5 moles NaOH x (1 mole Na2SO4 / 2 moles NaOH) = 0.25 moles
Step 3: Convert the moles of Na2SO4 to grams of Na2SO4.
The molar mass of Na2SO4 is:
2(23.0 g/mol) + 32.0 g/mol + 4(16.0 g/mol) = 142.0 g/mol
So, the theoretical yield of Na2SO4 is:
0.25 moles x 142.0 g/mol = 35.5 g
Step 4: Calculate the percent yield.
The actual mass of Na2SO4 collected is the total mass of the beaker and the dry compound minus the mass of the beaker itself.
Actual mass of Na2SO4 collected = 142.3 g - 128.4 g = 13.9 g
The percent yield is:
(13.9 g / 35.5 g) x 100 = 39.2%
Therefore, the percent yield of Na2SO4 is 39.2%.