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Chapter 9: Problem 15

(a) Balance this unbalanced equation by inspection: $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{H}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} $$ Benzene Hydrogen Cyclohexane (b) Express this reaction in words, using the word mole(s) wherever appropriate. (c) To produce 1 mole of \(\mathrm{C}_{6} \mathrm{H}_{12}\) from this reaction, how many grams of \(\mathrm{C}_{6} \mathrm{H}_{6}\) and \(\mathrm{H}_{2}\) must you combine? (d) What is the theoretical yield in grams of \(\mathrm{C}_{6} \mathrm{H}_{12} ?\) (e) Suppose you recover \(24.0 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{1}\). What is the percent vield?

Short Answer

Expert verified
(a) Balanced equation: \(C_6H_6 + 3H_2 \longrightarrow C_6H_{12}\) (b) 1 mole of benzene reacts with 3 moles of hydrogen gas to produce 1 mole of cyclohexane. (c) \(78.12 \: g\) of \(C_6H_6\) and \(6.06 \: g\) of \(H_2\) (d) Theoretical yield: \(84.18 \: g\) of \(C_6H_{12}\) (e) Percent Yield: \(28.52\%\)

Step by step solution

01

(a) Balance the Equation

To balance the given chemical equation, we have to adjust the coefficients of the reactants and products so that the number of atoms of each element is the same on both sides of the equation. Given equation: \(C_6H_6 + H_2 \longrightarrow C_6H_{12}\) Balanced equation: \(C_6H_6 + 3H_2 \longrightarrow C_6H_{12}\)
02

(b) Express the Reaction in Words

Now that the equation is balanced, we can express it in words: 1 mole of benzene reacts with 3 moles of hydrogen gas to produce 1 mole of cyclohexane.
03

(c) Calculate the Amount of Reactants

To determine the amounts of reactants required to produce 1 mole of \(C_6H_{12}\), we'll first find the molar mass of the reactants. Molar mass of \(C_6H_6 = (6 \times 12.01) + (6 \times 1.01) = 78.12 \: g/mol\) Molar mass of \(H_2 = 2 \times 1.01 = 2.02 \: g/mol\) From the balanced equation, 1 mole of \(C_6H_6\) and 3 moles of \(H_2\) are needed to produce 1 mole of \(C_6H_{12}\). Therefore, we need: \(78.12 \: g\) of \(C_6H_6\) and \(3 \times 2.02 \: g = 6.06 \: g\) of \(H_2\)
04

(d) Calculate the Theoretical Yield

The theoretical yield is the maximum amount of product that can be produced from the given reaction. To find the theoretical yield in grams of \(C_6H_{12}\), we first determine its molar mass. Molar mass of \(C_6H_{12} = (6 \times 12.01) + (12 \times 1.01) = 84.18 \: g/mol\) Since the balanced equation states that 1 mole of \(C_6H_6\) and 3 moles of \(H_2\) produce 1 mole of \(C_6H_{12}\), the theoretical yield is: \(1 \: mole \: of \: C_6H_{12} = 84.18 \: g\)
05

(e) Calculate the Percent Yield

To calculate the percent yield, we use the following formula: Percent Yield = \(\frac{\text{Actual Yield (g)}}{\text{Theoretical Yield (g)}} \times 100\% \) Actual yield = 24.0 grams Theoretical yield = 84.18 grams Percent Yield = \(\frac{24.0}{84.18} \times 100\% = 28.52\%\) So, the percent yield of the reaction is 28.52%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Understanding stoichiometry is fundamental to mastering chemical reactions. It's the study of the quantitative relationships or ratios between the reactants and products in a chemical reaction. It's like a recipe for chemistry where the coefficients of the compounds indicate the

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Most popular questions from this chapter

Lead(II) sulfide reacts with hydrogen peroxide to give lead(II) sulfate as shown in the unbalanced chemical equation \(\mathrm{PbS}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{PbSO}_{4}+\mathrm{H}_{2} \mathrm{O}\) If \(63.2 \mathrm{~g}\) of \(\mathrm{PbS}\) is reacted with \(48.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) (a) Which is the limiting reagent? (b) How many grams of excess reactant remain after the reaction is complete?

Consider the following unbalanced chemical equation: \(\mathrm{P}+\mathrm{O}_{2} \rightarrow \mathrm{P}_{2} \mathrm{O}_{5}\) (a) How many grams of phosphorus (P) are required to react completely with \(20.0 \mathrm{~g}\) of \(\mathrm{O}_{2} ?\) (b) What is the theoretical yield in grams if you combine the amounts of reactants in part (a)?

Consider the unbalanced chemical equation \(\mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{N}_{2} \mathrm{H}_{4} \rightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) Balance the equation. (b) If \(42.32 \mathrm{~g}\) of \(\mathrm{N}_{2}\) was produced and the reaction yield was \(67.5 \%\), how many grams of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{N}_{2} \mathrm{H}_{4}\) were consumed in the reaction?

The ingredients for making 1 dozen cupcakes are: 5 tablespoons butter 1 cup sugar 2 eggs 2 cups flour 1 cup milk If you have only 5 eggs but plenty of all of the other ingredients: (a) How many cupcakes can you make? (b) How many cups of sugar do you need to make the number of cupcakes calculated in (a)?

How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would you need to get \(5.00 \times 10^{30}\) carbon atoms?
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