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A student adds 4.00 g of dry ice (solid \(\mathrm{CO}_{2} )\) to an empty balloon. What will be the volume of the balloon at STP after all the dry ice sublimes (converts to gaseous \(\mathrm{CO}_{2} ) ?\)

Short Answer

Expert verified
After the 4.00 g of dry ice sublimes, the volume of the balloon at STP (Standard Temperature and Pressure) will be approximately 2.04 L.

Step by step solution

01

Calculate the moles of CO2

First, we need to find the number of moles of CO2 using the given mass and molar mass of CO2. Molar mass of CO2: C = 12.01 g/mol O = 16.00 g/mol Molar mass of CO2 = 1 × C + 2 × O = 1 × 12.01 g/mol + 2 × 16.00 g/mol = 44.01 g/mol Given mass of CO2: 4.00 g Now, we can calculate the number of moles (n) of CO2: n = (given mass) / (molar mass) n = 4.00 g / 44.01 g/mol ≈ 0.0909 mol
02

Calculate the volume of the balloon using Ideal Gas Law

Now that we have the number of moles of CO2 and the conditions are at STP, we can use the Ideal Gas Law equation to find the volume of the balloon. At STP, Temperature (T) = 273.15 K Pressure (P) = 1 atm Ideal Gas Law equation: \(PV=nRT\) We want to find the volume (V). Rearrange the equation for solving 'V': V = (nRT) / P Put the values into the equation and solve for the volume (V): V = (0.0909 mol × 0.0821 L atm/mol K × 273.15 K) / (1 atm) V = 2.0434 L So, after all the dry ice sublimes, the volume of the balloon at STP would be approximately 2.04 L.

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Key Concepts

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

Moles of Gas
Understanding the concept of "moles" is essential in chemistry, especially when dealing with gases and their behaviors. A mole is a unit that measures the amount of a substance. It's defined as exactly 6.022 x 10^23 particles, atoms, or molecules of that substance. It allows chemists to count particles by weighing them, which is particularly useful given the minuscule size of atoms and molecules.
For gases, the amount present in grams can be converted into moles using the substance's molar mass. The molar mass is the weight of one mole of a substance expressed in grams. For example, the molar mass of carbon dioxide (CO_{2}) is 44.01 grams per mole.
To find the moles of gas, use:
\[ = \frac{\text{given mass}}{\text{molar mass}}\]
This formula helps you calculate the number of moles from a known mass, which is critical for further calculations like volume or reacting masses in chemical equations.
STP Conditions
Understanding "STP conditions" is crucial when working with gas laws. STP stands for Standard Temperature and Pressure, which is a set of conditions often used as a reference point in calculations involving gases.
  • **Standard Temperature:** 273.15 Kelvin (K) - equivalent to 0 degrees Celsius (°C).
  • **Standard Pressure:** 1 atmosphere (atm) - a measure of the force exerted by the gas particles against the walls of a container.
These conditions are significant because they allow scientists to predict and compare the behavior of gases under a uniform set of parameters.
When gases are at these conditions, they exhibit predictable behavior as described by the Ideal Gas Law. Using STP conditions simplifies calculations because at STP, one mole of an ideal gas occupies exactly 22.4 liters, making it easier to determine the volume from the number of moles.
Volume Calculation
Volume calculation is a key aspect when using the Ideal Gas Law, which is represented as \(PV=nRT\). It's a formula that connects pressure (\(P\)), volume (\(V\)), number of moles (\(n\)), the gas constant (\(R\)), and temperature (\(T\)). To find volume at STP conditions, the equation is rearranged to solve for \(V\):
\[V = \frac{nRT}{P}\]
Where:
  • \(n\) is the number of moles (calculated from mass),
  • \(R\) is the ideal gas constant (0.0821 L atm/mol K),
  • \(T\) is the temperature in Kelvin, and
  • \(P\) is the pressure in atmospheres.
For example, when 4.00 grams of solid carbon dioxide sublime, using its molar mass to find \(n\), and substituting in STP values, you can determine the gas's volume. Solving \(V\) gives you the balloon's volume after the solid CO2 converts to gas at those specific conditions. Understanding this calculation allows predictions of how gases behave in different environmental settings.

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Most popular questions from this chapter

As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant pressure and temperature, the volume of the product gases collected is twice the volume of \(\mathrm{NH}_{3}\) reacted. Explain. As \(\mathrm{NH}_{3}(g)\) is decomposed into nitrogen gas and hydrogen gas at constant volume and temperature, the total pressure increases by some factor. Why the increase in pressure and by what factor does the total pressure increase when reactants are completely converted into products? How do the partial pressures of the product gases compare to each other and to the initial pressure of \(\mathrm{NH}_{3}?\)

Calculate the pressure exerted by 0.5000 mole of \(\mathrm{N}_{2}\) in a 10.000 -L container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results. d. Compare the results with those in Exercise 123.

Suppose two \(200.0-\mathrm{L}\) tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of 2.70 atm in its respective tank at \(24^{\circ} \mathrm{C} ?\)

You have a gas in a container fitted with a piston and you change one of the conditions of the gas such that a change takes place, as shown below: State three distinct changes you can make to accomplish this, and explain why each would work.

Metallic molybdenum can be produced from the mineral molybdenite, \(\mathrm{MoS}_{2}\). The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are $$\operatorname{MoS}_{2}(s)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow \operatorname{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g)$$ $$\mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)$$ Calculate the volumes of air and hydrogen gas at \(17^{\circ} \mathrm{C}\) and 1.00 atm that are necessary to produce \(1.00 \times 10^{3} \mathrm{kg}\) pure molybdenum from \(\mathrm{MoS}_{2}\) . Assume air contains 21\(\%\) oxygen by volume, and assume 100\(\%\) yield for each reaction.

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