Dichloromethane (CH2 Cl2), used as a solvent in the decaffeination of coffee beans, is prepared

Dichloromethane (CH2 Cl2), used as a solvent in the...

Dichloromethane $\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right),$ used as a solvent in the decaffeination of coffee beans, is prepared by reaction of methane (CH_) with chlorine. How many grams of dichloromethane result from reaction of 1.85 $\mathrm{kg}$ of methane if the yield is 43.1$\% ?$ $$\mathrm{CH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}(l)+2 \mathrm{HCI}(g)$$

Dichloromethane $\left(\mathrm{CH}_{2} \mathrm{Cl}_{2}\right),$ used as a solvent in the decaffeination of coffee beans, is prepared by reaction of methane (CH_) with chlorine. How many grams of dichloromethane result from reaction of 1.85 $\mathrm{kg}$ of methane if the yield is 43.1$\% ?$
$$\mathrm{CH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}(l)+2 \mathrm{HCI}(g)$$

Key Concepts

Stoichiometry

Stoichiometry involves using the balanced chemical equation to relate the amounts (in moles) of reactants to products. It is used to calculate how many moles of a certain product can be theoretically produced from a given amount of a reactant.

Molar Mass and Unit Conversion

Molar mass is the mass of one mole of a substance and is used as a conversion factor between the mass of a substance and the number of moles. This conversion is essential for relating laboratory-scale measurements (grams) to quantities needed in stoichiometric calculations (moles).

Percent Yield

Percent yield is a measure of the efficiency of a reaction, defined as the ratio of the actual yield to the theoretical yield, multiplied by 100%. It accounts for losses and incomplete reactions, providing a realistic amount of product expected from a reaction.

Reaction Yield Calculation

Reaction yield calculation combines stoichiometry and percent yield concepts to determine the actual amount of product formed. It involves computing the theoretical mass through stoichiometric conversion from reactant mass and then adjusting this value by the percent yield to account for practical limitations.

Transcript

00:01 In this question, it will serve us to have a formula for percent yield.

00:07 I use this equation to organize the information that i am given.

00:12 I am told that 1 .85 grams of methane are used.

00:19 So that's 1 .85 not grams, kilograms of methane.

00:27 I'm also told that the yield of dichloromethan is 43 .1%.

00:39 And i'm asked how many grams resulted from the reaction.

00:44 Now when they say that we have 43 .1 % yield and we're asked how much resulted, that's how much was physically made in the lab.

00:54 When i ran this reaction, how much did i actually isolate? so this is our goal.

01:00 This is what we're looking for.

01:03 So notice that in this equation of percent yield, i have two, sorry, excuse me, i have one of the three variables.

01:11 Goals, what i need to find first is the theoretical yield.

01:18 So this is my first step.

01:22 And once i have the theoretical yield, i'll be able to find the actual yield, which is my goal.

01:30 So to find the theoretical yield, i need to start with my reactant and use it to calculate the theoretical yield, which is if all of this 1 .85 kilograms reacts and makes perfectly all the dichloromethine possible, how much dichloromethine will that be? so let's start.

01:58 First by recognizing that 1 .85 kilograms is equal to 1 ,850 grams.

02:07 We need grams because our molar masses are ingrained.

02:10 So we start with 1850 grams of methane.

02:17 We divide by its molar mass, which is 16 .04 grams of methane for every one mole of methane...

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