1.Chlorine gas can be produced commercially
by passing an electric current through a concentrated
solution of sodium chloride (brine).
The unbalanced equation is
NaCl(aq) + H2O(ℓ) →
NaOH(aq) + Cl2(g) + H2(g).
If the brine contains 163 g of NaCl, how
much Cl2 can be produced?
Answer in units of mol.
This is what I did
(163 g NaCl) / (58.4430 g NaCl/mol) x (1/2) = 1.39 mol Cl2
(Part 2 of 2)How much H2 can be produced?
Answer in units of mol.
I don't know how to do the second questions since they didn't give how much h2 weighs
Please help me set up the problem
2.Phosphorous trichloride (PCl3) is produced
from the reaction of white phosphorous (P4)
and chlorine:
P4(s) + 6 Cl2(g) → 4 PCl3(g).
A sample of PCl3 of mass 300.1 g was collected
from the reaction of 72.55 g of P4 with excess
chlorine. What is the percentage yield of the
reaction?
Answer in units of %.
To calculate the amount of H2 produced in the reaction in Part 1 of the question, we need to use the stoichiometry of the balanced equation.
The balanced equation is: NaCl(aq) + H2O(ℓ) → NaOH(aq) + Cl2(g) + H2(g)
From the equation, we can see that the mole ratio of NaCl to Cl2 is 1:1 and the mole ratio of NaCl to H2 is 1:1.
First, let's calculate the amount of Cl2 produced:
Using the given mass of NaCl (163 g) and the molar mass of NaCl (58.4430 g/mol), we can find the number of moles of NaCl:
(163 g NaCl) / (58.4430 g NaCl/mol) = 2.7909 mol NaCl
Since the mole ratio of NaCl to Cl2 is 1:1, the amount of Cl2 produced is also 2.7909 mol.
Now, let's calculate the amount of H2 produced:
Since the mole ratio of NaCl to H2 is 1:1, the amount of H2 produced is also 2.7909 mol.
Therefore, the answer to the second part of the question is 2.7909 mol of H2.
Moving on to the second question:
To calculate the percentage yield of the reaction in Part 2, we need to compare the actual yield (the mass of PCl3 collected) to the theoretical yield (the maximum possible mass of PCl3 that can be obtained based on the balanced equation).
From the balanced equation: P4(s) + 6 Cl2(g) → 4 PCl3(g)
We can see that the mole ratio of P4 to PCl3 is 1:4.
First, let's calculate the theoretical yield of PCl3:
Using the given mass of P4 (72.55 g) and the molar mass of PCl3 (137.33 g/mol), we can find the number of moles of P4:
(72.55 g P4) / (123.8955 g P4/mol) = 0.5855 mol P4
Since the mole ratio of P4 to PCl3 is 1:4, the theoretical yield of PCl3 is:
0.5855 mol P4 x (4 mol PCl3 / 1 mol P4) = 2.342 mol PCl3
The molar mass of PCl3 is 137.33 g/mol, so the theoretical mass of PCl3 is:
2.342 mol PCl3 x (137.33 g/mol) = 321.39 g PCl3
Now, let's calculate the percentage yield:
The actual yield of PCl3 is given as 300.1 g.
Percentage yield = (actual yield / theoretical yield) x 100
= (300.1 g / 321.39 g) x 100
= 93.38 %
Therefore, the answer to the second part of the question is 93.38 %.
To solve the second question, you can use the balanced equation and the molar mass of PCl3 to find the amount of PCl3 produced.
The balanced equation is:
P4(s) + 6 Cl2(g) → 4 PCl3(g)
To find the molar mass of PCl3, you can add up the atomic masses of each element:
P (phosphorous) = 30.97 g/mol
Cl (chlorine) = 35.45 g/mol
Molar mass of PCl3 = (30.97 g/mol) + 3 * (35.45 g/mol) = 137.33 g/mol
Using the molar mass of PCl3, you can now calculate the moles of PCl3 produced:
Moles of PCl3 = mass of PCl3 / molar mass of PCl3
Moles of PCl3 = 300.1 g / 137.33 g/mol = 2.185 mol PCl3
From the balanced equation, you can see that the stoichiometric ratio between P4 and PCl3 is 1:4. This means that for every mole of P4, you get 4 moles of PCl3.
Therefore, the number of moles of P4 used in the reaction can be calculated as:
Moles of P4 = 1/4 * Moles of PCl3
Moles of P4 = 1/4 * 2.185 mol = 0.546 mol P4
Now, you can calculate the theoretical yield of PCl3, which is the maximum amount of PCl3 that can be produced from the given amount of P4:
Theoretical yield of PCl3 = 4 * Moles of P4
Theoretical yield of PCl3 = 4 * 0.546 mol = 2.184 mol PCl3
Finally, you can calculate the percentage yield by dividing the actual yield (given in the question) by the theoretical yield, and multiplying by 100:
Percentage yield = (Actual yield / Theoretical yield) * 100
Percentage yield = (2.185 mol / 2.184 mol) * 100 = 100.05% (rounded to two decimal places)
Therefore, the percentage yield of the reaction is approximately 100.05%.
I think for part 1 they want grams. Your 1.39 mols Cl2 is right; just multiply by molar mass Cl2.
part 2. H2 done the same way you did #1.
mols NaCl = 163/58.44 = ?
mols H2 = 1/2 that from the balanced equation.
grams H2 = mols H2 x molar mass H2 which is 2 g/mol.
#2. P4(s) + 6 Cl2(g) → 4 PCl3(g).
mols P4 = 72.55/molar mass P4 = ?
Convert mols P4 to mols PCl3 using the coefficients in the balanced equation. That's ?mols P4 x (4 mols PCl3/1 mol P4) = ? mols PCl3.
Then g PCl3 = mols PCl3 x molar mass PCl3. This is the theoretical yield(TY); ie., as if it were 100%. The actual yield (AY) is stated in the problem as 72.55.
Then % yield = (AY/TY)*100 = ?