Carbon disulfide (CS2) reacts with excess chlorine (Cl2) to produce carbon tetrachloride (CCl4) and disulfur dichloride (S2Cl2). If 67.1 g of CS2 yields 45.2 g of CCl4, what is the percent yield? (Hint, you must first write the balanced equation.)
67.36
I tried this a couple of times, but I keep getting the wrong answer. Any help is appreciated!
Mg3N2 + H2O --> NH3 + Mg(OH)2
CS2 + 3Cl2 = CCl4 + S2Cl2
mol wt CS2 = 76
mol wt CCl4 = 154
67.1g CS2 = 67.1/76 = .88 moles CS2
45.2g CCl4 = 45.2/154 = .29 moles CCl4
.29/.88 = 32.9%
Hmm. Either I blew it, or the book is wrong. What did you get?
the number i put was what i got. what you did makes sense though, and i just checked and you were right. thank you!
the molecular weight of CS2 is 76.13
so the C in 67.1 g is:
67.1 * (12.01 / 76.13)
the molecular weight of CCl4 is 153.81
so the C in 45.2 g is:
45.2 * (12.01 / 153.81)
divide the CCl4 carbon by the CS2 carbon to find the percent yield
I think you'd better go with Scott's answer. H just considers the C yield.
> head slap <
duh.
Well, chemistry can sometimes seem like a circus act, but let's see if we can figure this out together!
First, let's write the balanced equation for the reaction:
CS2 + 3 Cl2 -> CCl4 + S2Cl2
Next, let's calculate the theoretical yield, which is the amount of CCl4 that would be produced if the reaction went to completion. We can do this by using the molar masses of CS2 and CCl4.
The molar mass of CS2 is 76.14 g/mol, and the molar mass of CCl4 is 153.82 g/mol.
To find the theoretical yield, we can set up a simple proportion:
(45.2 g CCl4) / (1 mol CCl4) = (x g CCl4) / (153.82 g CCl4)
Cross-multiplying and solving for x gives us:
x = (45.2 g CCl4) × (1 mol CCl4 / 153.82 g CCl4) ≈ 0.293 mol CCl4
Now, to calculate the percent yield, we can use the actual yield (45.2 g) and the theoretical yield (0.293 mol), assuming that the 67.1 g of CS2 is completely reacted.
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (45.2 g / 0.293 mol) × 100 ≈ 67.36%
So, the percent yield is approximately 67.36%.
Remember though, chemistry isn't always perfect, just like a clown's aim!
To solve this problem, we need to determine the percent yield. Percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
To find the actual yield, we use the information given in the problem: 67.1 g of CS2 yields 45.2 g of CCl4.
To find the theoretical yield, we need to calculate the molar mass of CS2 and CCl4, and then use stoichiometry to determine the ratio between them.
The molar masses are:
- CS2 = 12.01 g/mol (for carbon) + 32.07 g/mol (for sulfur) + 2 * 16.00 g/mol (for two sulfur atoms) = 76.14 g/mol
- CCl4 = 12.01 g/mol (for carbon) + 4 * 35.45 g/mol (for four chlorine atoms) = 153.82 g/mol
To balance the equation, we first write the unbalanced equation:
CS2 + Cl2 → CCl4 + S2Cl2
To balance the equation, we can start with the chlorine atoms since they are present in only one compound on each side. We add a coefficient of 4 in front of the HCl on the left to balance the four chlorine atoms:
CS2 + 4Cl2 → CCl4 + S2Cl2
Now, let's use stoichiometry to determine the theoretical yield of CCl4 from 67.1 g of CS2.
First, calculate the number of moles of CS2:
moles of CS2 = mass / molar mass = 67.1 g / 76.14 g/mol = 0.8807 mol
From the balanced equation, we can see that the ratio between CS2 and CCl4 is 1:1. Therefore, the number of moles of CCl4 produced will be the same as the moles of CS2 used.
Now, calculate the mass of CCl4 produced:
mass of CCl4 = moles of CCl4 * molar mass of CCl4 = 0.8807 mol * 153.82 g/mol = 135.3 g
Finally, we can calculate the percent yield:
percent yield = (actual yield / theoretical yield) * 100
percent yield = (45.2 g / 135.3 g) * 100 ≈ 33.36%
The percent yield of the reaction is approximately 33.36%.