Question

For the following reaction, 7.31 grams of methane (CH4) are allowed to react with 29.2 grams of carbon tetrachloride . methane (CH4) ( g ) + carbon tetrachloride ( g ) dichloromethane (CH2Cl2) ( g ) what amount of the excess reagent remains after the reaction is complete?

          For the following reaction, 7.31 grams of methane (CH4) are allowed to react with 29.2 grams of carbon tetrachloride .
methane (CH4) ( g ) + carbon tetrachloride ( g )  dichloromethane (CH2Cl2) ( g )
what amount of the excess reagent remains after the reaction is complete?

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Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Chareen Guzman

05/05/2023

Step 1

Moles of methane (CH4) = mass / molar mass = 7.31 g / (12.01 g/mol + 4 * 1.01 g/mol) = 7.31 g / 16.05 g/mol = 0.455 moles Moles of carbon tetrachloride (CCl4) = mass / molar mass = 29.2 g / (12.01 g/mol + 4 * 35.45 g/mol) = 29.2 g / 153.81 g/mol = 0.190 moles Show more…

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For the following reaction, 7.31 grams of methane (CH4) are allowed to react with 29.2 grams of carbon tetrachloride . methane (CH4) ( g ) + carbon tetrachloride ( g ) dichloromethane (CH2Cl2) ( g ) what amount of the excess reagent remains after the reaction is complete?

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00:01 Hi, so to solve for this, let us write first the balance equation for the reaction of methane and carbon tetrachloride.

00:10 So that's ch4, this is methane, and carbon tetrachloride is ccl4.

00:16 Then the product, we have ch2cl2.

00:21 So to balance this, we just need to write here 2.

00:25 Okay, so this is the balance equation.

00:27 Now we need to identify the limiting reactant and the excess reactant in this reaction.

00:34 So we'll solve first for the number of moles of each reactant.

00:38 We'll start with methane, number of moles of ch4.

00:42 If we have 7 .31, this is grams of methane.

00:47 Convert this to moles by dividing the molar mass of ch4.

00:50 That's 16 .05 grams here, and then moles in the numerator.

00:56 So that means we could cancel grams of ch4 and we'll get 0 .455.

01:04 This is moles of ch4.

01:07 And then for carbon tetrachloride, we have 29 .2 grams of ccl4.

01:17 Convert to moles by dividing the molar mass of ccl4.

01:21 That's going to be 153 .81 grams here, and then moles in the numerator.

01:27 So that means we could cancel grams and we'll get 0 .190.

01:34 This is moles of ccl4.

01:37 Now we will solve for the mole ratio.

01:42 So mole ratio in this case is the number of moles of the reactant divided by the coefficient of that reactant from the balance equation.

01:50 So for ch4, we have calculated 0 .455 moles divided by the coefficient of ch4 from the balance equation, which is 1.

02:02 So divided by 1, we'll still get 0 .455.

02:07 And then for ccl4, we have 0 .190.

02:13 This is moles divided by the coefficient of carbon tetrachloride, which is also 1...

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