Carbon disulfide is produced by the reaction of carbon and...

Carbon disulfide is produced by the reaction of carbon and sulfur dioxide. $$ 5 \mathrm{C}(s)+2 \mathrm{SO}_{2}(g) \longrightarrow \mathrm{CS}_{2}(g)+4 \mathrm{CO}(g) $$ a. What is the percent yield of carbon disulfide if the reaction of $40.0 \mathrm{~g}$ of carbon produces $36.0 \mathrm{~g}$ of carbon disulfide? b. What is the percent yield of carbon disulfide if the reaction of $32.0 \mathrm{~g}$ of sulfur dioxide produces $12.0 \mathrm{~g}$ of carbon disulfide?

Carbon disulfide is produced by the reaction of carbon and sulfur dioxide.
$$
5 \mathrm{C}(s)+2 \mathrm{SO}_{2}(g) \longrightarrow \mathrm{CS}_{2}(g)+4 \mathrm{CO}(g)
$$
a. What is the percent yield of carbon disulfide if the reaction of $40.0 \mathrm{~g}$ of carbon produces $36.0 \mathrm{~g}$ of carbon disulfide?
b. What is the percent yield of carbon disulfide if the reaction of $32.0 \mathrm{~g}$ of sulfur dioxide produces $12.0 \mathrm{~g}$ of carbon disulfide?

Key Concepts

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. By using the balanced chemical equation, one can determine the mole ratios of substances and compute how much product can theoretically be formed from given amounts of reactants.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is completely consumed first, thereby restricting the amount of product that can be formed. Identifying the limiting reactant is key to calculating the theoretical yield of a reaction.

Theoretical Yield

The theoretical yield is the maximum possible amount of product that can be formed from a chemical reaction, calculated based on the stoichiometry of the reaction and assuming complete conversion of the limiting reactant.

Percent Yield

Percent yield compares the actual yield, which is the measured amount of product obtained from an experiment, to the theoretical yield. It is expressed as a percentage and serves as a measure of the efficiency of a chemical reaction.

Transcript

00:01 Okay, so for this problem, this is similar to others that are in your textbook.

00:07 So this is all about theoretical actual yield and percent yield.

00:12 So as a reminder, percent yield equals actual yield, which will be given to you, divided by theoretical yield, which will be calculated prior to the lab, hopefully.

00:33 This would be what happens if everything goes perfectly well, and you multiply that value by 100 to get the percent.

00:40 So for part a, we're starting with 40 grams of carbon, and we're trying to figure out how much cs2 can be produced because our actual yield is 36 grams of cs2.

01:01 So to start, we've got to take what we know, which is 40 .0 grams of carbon.

01:09 My blue values are all coming from the periodic table.

01:12 So that's the mass off the periodic table, which is 12 .01.

01:18 And that is equivalent to one mole of that substance.

01:22 Then we have to get rid of moles.

01:24 So we put moles on the bottom.

01:26 And we're trying to get to grams of cs2, excuse me.

01:30 So we're going to put mole of cs2.

01:32 Because we can't go straight two grams of an unknown substance.

01:35 So when we have mole on top of mole, we have to look at our coefficients.

01:39 For carbon, it is five for cs.

01:42 It is 1.