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Chapter 12: Problem 52

Consider the following equation: $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) How many liters of oxygen are required to react with \(7.2 \mathrm{~L}\) \(\mathrm{C}_{3} \mathrm{H}_{8}\) ? Both gases are at STP. (b) How many grams of \(\mathrm{CO}_{2}\) will be produced from \(35 \mathrm{~L} \mathrm{C}_{3} \mathrm{H}_{8}\) if both gases are at STP? (c) How many liters of water vapor can be produced when \(15 \mathrm{~L}\) \(\mathrm{C}_{3} \mathrm{H}_{8}\) are reacted with \(15 \mathrm{~L} \mathrm{O}_{2}\) ? All gases are at the same temperature and pressure.

Short Answer

Expert verified
Part (a): 36 L O_2Part (b): 206.25 g CO_2 Part (c): 6 L H_2O

Step by step solution

01

Understand the Stoichiometry

First, observe the balanced equation: \[ \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \] This tells us the ratio in which the reactants and products react and are produced.
02

Calculate Required Oxygen (Part a)

Use the stoichiometric ratios given. For every 1 volume of \text{C}_3\text{H}_8, 5 volumes of \text{O}_2 are needed. Given: 7.2 L \text{C}_3\text{H}_8 Required: \[ 7.2 \text{L} \text{C}_3\text{H}_8 \times \frac{5 \text{L} \text{O}_2}{1 \text{L} \text{C}_3\text{H}_8} = 36 \text{L} \text{O}_2 \]
03

Calculate Produced Carbon Dioxide (Part b)

Given 35 L \text{C}_3\text{H}_8, we need to find the grams of \text{CO}_2 produced. 1 volume of \text{C}_3\text{H}_8 produces 3 volumes of \text{CO}_2.Calculate volume of \text{CO}_2: \[ 35 \text{L} \text{C}_3\text{H}_8 \times \frac{3 \text{L} \text{CO}_2}{1 \text{L} \text{C}_3\text{H}_8} = 105 \text{L} \text{CO}_2 \] At STP, 1 mole of gas occupies 22.4 L. Calculate moles of \text{CO}_2: \[ \frac{105 \text{L}}{22.4 \text{L/mol}} = 4.6875 \text{mol} \text{CO}_2 \] Molar mass of \text{CO}_2 = 44 g/mol, hence mass of \text{CO}_2: \[ 4.6875 \text{mol} \times 44 \text{g/mol} = 206.25 \text{g} \]
04

Calculate Produced Water Vapor (Part c)

For every 1 volume of \text{C}_3\text{H}_8, 4 volumes of \text{H}_2\text{O} are produced. Given: 15 L \text{C}_3\text{H}_8 and 15 L \text{O}_2 Limiting reagent is \text{O}_2 as it must be in a ratio of 1:5. The 15 L \text{O}_2 satisfies the amount of \text{C}_3\text{H}_8 available as \[ 15 L \text{C}_3\text{H}_8 \times \frac{5 L \text{O}_2}{1 L \text{C}_3\text{H}_8} = 75 L \text{O}_2 which is more than available for 15 L \text{O}_2, making \text{O}_2 limiting, hence the available quantity of \text{C}_3\text{H}_8 will be limited by \text{O}_2 to \text{15/5} = 3 \text{L}giving an effective product with 3 \text{L} of only the available reactant entity. \] The produced water vapor: \[ 3 \text{L} \text{C}_3\text{H}_8 \times \frac{4 \text{L} \text{H}_2\text{O}}{1 \text{L} \text{C}_3\text{H}_8=\text{12} ) = 12 L \text{H}_2\text{O} \]

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Key Concepts

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

Chemical Reactions
A chemical reaction is when substances combine or break apart to form new substances. In our example, propane (C3H8) reacts with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation is: \[ \text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \] This balanced equation tells us the ratio in which the reactants combine and products are formed. For instance, 1 molecule of propane reacts with 5 molecules of oxygen to produce 3 molecules of carbon dioxide and 4 molecules of water. Understanding this ratio is crucial to solving stoichiometry problems, as it allows us to predict quantities of reactants and products.
Gas Laws
Gas laws describe the behavior of gases under different conditions of temperature, pressure, and volume. One of the most important laws in stoichiometry of gases is the Ideal Gas Law, represented by \( PV = nRT \), where:
  • P is the pressure
  • V is the volume
  • n is the number of moles
  • R is the ideal gas constant
  • T is the temperature
In stoichiometry problems, we often work at STP (Standard Temperature and Pressure), which is 0°C (273 K) and 1 atm pressure. At STP, 1 mole of any gas occupies 22.4 liters. This simplifies many calculations, like finding the volume of gases when the number of moles is known, or vice versa. For instance, in our exercise, since both gases are at STP, we can use the molar volume directly to find volumes and convert between liters and moles easily.
Mole Conversions
Mole conversions allow us to switch between mass, volume, and the number of particles using proportions. For example, to convert volumes of gases at STP to moles, we use the fact that 1 mole of gas occupies 22.4 liters. In part (b) of our exercise, we converted the volume of CO2 produced from propane to grams. Steps for mole conversions:
  • Convert volume to moles using 22.4 L/mol at STP: \( \text{moles} = \frac{\text{volume}}{22.4} \)
  • Use the molar mass to convert moles to grams: \( \text{mass} = \text{moles} \times \text{molar mass} \)
For example, in step 2 of our exercise:
Given 105 L CO2 at STP: \[ \text{moles of CO2} = \frac{105 \text{L}}{22.4 \text{L/mol}} = 4.6875 \text{mol} \] Then, convert moles of CO2 to grams: \[ \text{mass} = 4.6875 \text{mol} \times 44 \text{g/mol} = 206.25 \text{g} \] By understanding and practicing these conversions, solving stoichiometry problems becomes more straightforward.

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Most popular questions from this chapter

$$ \begin{aligned} &\text { Fill in the table below with the missing information: }\\\ &\begin{array}{|c|c|c|c|} \hline & \text { torr } & \text { in. Hg } & \text { kilopascals (kPa) } \\ \hline \text { (a) } & & 30.2 & \\ \hline \text { (b) } & 752 & & \\ \hline \text { (c) } & & & 99.3 \\ \hline \end{array} \end{aligned} $$

(a) If you filled up three balloons with equal volumes of hydrogen, argon, and carbon dioxide gas, all at the same temperature and pressure, which balloon would weigh the most? the least? Explain. (b) If you filled up three balloons with equal masses of nitrogen, oxygen, and neon, all to the same volume at the same temperature, which would have the lowest pressure?

A sample of \(\mathrm{CH}_{4}\) gas occupies a volume of \(575 \mathrm{~mL}\) at \(-25^{\circ} \mathrm{C}\). If pressure remains constant, what will be the new volume if temperature changes to: (a) \(298 \mathrm{~K}\) (b) \(32^{\circ} \mathrm{F}\) (c) \(45^{\circ} \mathrm{C}\)

A gas has a density at STP of \(1.78 \mathrm{~g} / \mathrm{L}\). What is its molar mass?

A sample of \(\mathrm{CO}_{2}\) gas occupies a volume of \(125 \mathrm{~mL}\) at \(21^{\circ} \mathrm{C}\). If pressure remains constant, what will be the new volume if temperature changes to: (a) \(-5^{\circ} \mathrm{C}\) (b) \(95^{\circ} \mathrm{F}\) (c) \(1095 \mathrm{~K}\)
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