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Chapter 5: Problem 40

Calculate its volume (in liters) of \(88.4 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) at STP.

Short Answer

Expert verified
The volume of 88.4 g of \(\mathrm{CO}_{2}\) at STP is 44.8 L.

Step by step solution

01

Determining the molar mass

First, we need to determine the molar mass of carbon dioxide (\( \mathrm{CO}_{2}\)), in terms of grams per mole (\( g/mol \)). Carbon has a molar mass of 12.01 g/mol and oxygen has a molar mass of 16.00 g/mol. Since there are two oxygen atoms for every one carbon atom in \( \mathrm{CO}_{2}\), the molar mass \( M \) is \( 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol \).
02

Calculating number of moles

Next, divide the mass given in the problem by the molar mass of \( \mathrm{CO}_{2}\) to find the number of moles \( n \). We know the mass \( m \) is 88.4 g and the molar mass \( M \) is 44.01 g/mol, so the number of moles \( n \) is \( m / M = 88.4 \, g / 44.01 \, g/mol = 2 \, moles \).
03

Calculating volume

Finally, multiply the number of moles of \( \mathrm{CO}_{2}\) by the volume that one mole of any gas occupies at STP. We know that one mole of any gas occupies a volume of 22.4 liters at STP. So, the volume \( V \) is \( n * 22.4 \, L/mol = 2 \, moles * 22.4 \, L/mol = 44.8 \, L \). So, 88.4 g of \( \mathrm{CO}_{2}\) at STP occupies a volume of 44.8 liters.

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Key Concepts

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

Molar Mass
Understanding molar mass is essential when dealing with gas laws. The molar mass of a compound is the mass of one mole of that substance, typically expressed in grams per mole (g/mol). For carbon dioxide (\(\mathrm{CO}_2\)), the molar mass is calculated by adding the atomic masses of its constituent elements. Carbon has a molar mass of 12.01 g/mol and oxygen is 16.00 g/mol. Since there are two oxygen atoms, you calculate \(12.01 + 2 \times 16.00 = 44.01 \ g/mol\) for \(\mathrm{CO}_2\). Knowing the molar mass helps in converting between the mass of a substance and the number of moles.
Standard Temperature and Pressure (STP)
Standard Temperature and Pressure, or STP, is a set of conditions used as a reference point in gas calculations. At STP, the temperature is 0掳C (273.15 K) and the pressure is 1 atmosphere (atm). This standardization aids in comparing different gas samples and simplifies calculations. At STP, one mole of any ideal gas occupies 22.4 liters. This concept is crucial for converting between moles and volume in gas law problems.
Mole Calculation
Moles are a fundamental unit in chemistry representing a specific amount of substance. To determine the number of moles from a given mass, divide the mass by the molar mass of the compound. In this exercise, the mass of \(\mathrm{CO}_2\) is 88.4 g. By using the molar mass \(44.01 \ g/mol\), the calculation is \(88.4 \ g \/ 44.01 \ g/mol = 2\ moles\). Understanding mole conversions ensures accurate calculations for reactions and gas volumes.
Volume of Gas at STP
The volume of a gas at STP is determined using the relationship where one mole of gas occupies 22.4 liters. For the problem at hand, after calculating that there are 2 moles of \(\mathrm{CO}_2\), you multiply the moles by the volume per mole at STP: \(2 \times 22.4 \ L/mol = 44.8 \ L\). This demonstrates how moles can be directly translated into a volume measurement using standard conditions.

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Most popular questions from this chapter

A quantity of \(0.225 \mathrm{~g}\) of a metal \(\mathrm{M}\) (molar mass = \(27.0 \mathrm{~g} / \mathrm{mol}\) ) liberated \(0.303 \mathrm{~L}\) of molecular hydrogen (measured at \(17^{\circ} \mathrm{C}\) and \(741 \mathrm{mmHg}\) ) from an excess of hydrochloric acid. Deduce from these data the corresponding equation and write formulas for the oxide and sulfate of M.

Why is the density of a gas much lower than that of a liquid or solid under atmospheric conditions? What units are normally used to express the density of gases?

(a) Show that the pressure exerted by a fluid \(P\) (in pascals) is given by \(P=h d g,\) where \(h\) is the column of the fluid in meters, \(d\) is the density in \(\mathrm{kg} / \mathrm{m}^{3},\) and \(g\) is the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) (Hint: See Appendix 1.) (b) The volume of an air bubble that starts at the bottom of a lake at \(5.24^{\circ} \mathrm{C}\) increases by a factor of 6 as it rises to the surface of water where the temperature is \(18.73^{\circ} \mathrm{C}\) and the air pressure is 0.973 atm. The density of the lake water is \(1.02 \mathrm{~g} / \mathrm{cm}^{3}\). Use the equation in (a) to determine the depth of the lake in meters.

The shells of hard-boiled eggs sometimes crack due to the rapid thermal expansion of the shells at high temperatures. Suggest another reason why the shells may crack.

The temperature of \(2.5 \mathrm{~L}\) of a gas initially at \(\mathrm{STP}\) is raised to \(250^{\circ} \mathrm{C}\) at constant volume. Calculate the final pressure of the gas in atm.
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