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Chapter 9: Problem 24

In the burning of propane, $$ \mathrm{C}_{3} \mathrm{H}_{8}+5 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O} $$ suppose \(10.0 \mathrm{~g}\) of propane is combined with \(10.0 \mathrm{~g}\) of oxygen. What is the theoretical yield of \(\mathrm{CO}_{2}\), in grams? Answer: Before we begin, we'll calculate the molar masses of everything because we'll probably need them: \(\mathrm{C}_{3} \mathrm{H}_{8}=44.096 \mathrm{~g} / \mathrm{mol} \quad \mathrm{O}_{2}=31.998 \mathrm{~g} / \mathrm{mol}\) \(\mathrm{CO}_{2}=44.009 \mathrm{~g} / \mathrm{mol} \quad \mathrm{H}_{2} \mathrm{O}=18.015 \mathrm{~g} / \mathrm{mol}\)

Short Answer

Expert verified
The theoretical yield of CO₂ is \(1.65 \thinspace g\).

Step by step solution

01

Convert the mass of propane and oxygen to moles

Using the provided molar masses, we will convert the mass of propane and oxygen to moles: -For propane: \( \dfrac{10.0 \thinspace g \thinspace propane}{44.096 \thinspace g/mol \thinspace propane} = 0.2267 \thinspace mol \thinspace propane \) -For oxygen: \( \dfrac{10.0 \thinspace g \thinspace oxygen}{31.998 \thinspace g/mol \thinspace oxygen} = 0.3125 \thinspace mol \thinspace oxygen \)
02

Determine the limiting reactant

To figure out which reactant is the limiting reactant, we will divide the moles of each reactant by their respective stoichiometric coefficients from the balanced equation: - For propane: \( \dfrac{0.2267 \thinspace mol \thinspace propane}{1} = 0.2267 \) - For oxygen: \( \dfrac{0.3125 \thinspace mol \thinspace oxygen}{5} = 0.0625 \) Since the value for oxygen is smaller (0.0625), oxygen is the limiting reactant in this chemical reaction.
03

Calculate the theoretical yield of CO₂

Now, using the stoichiometry of the balanced equation and the moles of the limiting reactant (oxygen), we can calculate the moles of CO₂ produced: \(0.0625 \thinspace mol \thinspace O_2 \times \frac{3 \thinspace mol \thinspace CO_2}{5 \thinspace mol \thinspace O_2} = 0.0375 \thinspace mol \thinspace CO_2 \) To find the theoretical yield of CO₂ in grams, multiply the moles of CO₂ by its molar mass: \(0.0375 \thinspace mol \thinspace CO₂ \times \dfrac{44.009 \thinspace g}{mol \thinspace CO₂} = 1.65 \thinspace g \thinspace CO₂ \) Therefore, the theoretical yield of CO₂ is 1.65 grams.

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Most popular questions from this chapter

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