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Consider the following reaction: $$ 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) $$ It takes \(2.00 \mathrm{~L}\) of pure oxygen gas at STP to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

Short Answer

Expert verified
The mass of aluminum reacted in the reaction is approximately 3.21 g.

Step by step solution

01

Convert the volume of oxygen gas to moles

At STP (standard temperature and pressure), any gas occupies 22.4 liters per mole. We have the volume of oxygen gas which is 2.00 L. To find the number of moles of oxygen, we'll use the following formula: \(n = \frac{V}{V_m}\) Where: \(n\) = number of moles, \(V\) = volume of the gas (2.00 L), \(V_m\) = molar volume of the gas at STP (22.4 L/mol). \(n = \frac{2.00}{22.4}\) Calculate the moles of oxygen: \( n = 0.0893 \, \text{mol} \)
02

Determine the moles of aluminum reacted

Now we have the moles of oxygen gas, we can use the stoichiometry of the balanced chemical equation to find the moles of aluminum reacted. The balanced chemical equation is: \(4 \, \text{Al}(s) + 3 \, \text{O}_{2}(g) \longrightarrow 2 \, \text{Al}_{2}\text{O}_{3}(s)\) The stoichiometric ratio of aluminum to oxygen in the balanced equation is 4:3. To find the moles of aluminum reacted, we simply set up a proportion: \(\frac{\text{moles of Al}}{\text{moles of O}_{2}} = \frac{4}{3}\) We know the moles of oxygen gas is 0.0893 mol, so we can solve for the moles of aluminum: \(\text{moles of Al} = \frac{4}{3} \times 0.0893\) Calculate the moles of aluminum: \(\text{moles of Al} = 0.119 \, \text{mol} \)
03

Convert moles of aluminum to mass

Now that we have the moles of aluminum, we can convert it to mass using the molar mass of aluminum. The molar mass of aluminum is approximately 26.98 g/mol. To find the mass of aluminum reacted, we'll use the following conversion: mass of aluminum = moles of aluminum × molar mass of aluminum mass of aluminum = 0.119 mol × 26.98 g/mol Calculate the mass of aluminum: mass of aluminum = 3.21 g The mass of aluminum reacted in the reaction is approximately 3.21 g.

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Key Concepts

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

Molar Volume of Gas
When dealing with gases, it is crucial to understand the concept of molar volume. At Standard Temperature and Pressure (STP), which is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atmosphere, one mole of any ideal gas occupies 22.4 liters.

This relationship acts as a bridge between the microscopic world of molecules and the macroscopic world we can measure. When the problem stated that 2.00 L of oxygen gas reacted at STP, it provided a direct way to calculate the number of moles of oxygen by utilizing the molar volume.

The formula \(n = \frac{V}{V_m}\) is used, where \(n\) is the number of moles, \(V\) is the actual volume of the gas we have, and \(V_m\) is the molar volume. Regardless of the type of gas, this conversion is possible due to the universal nature of the molar volume of gases at STP.
Chemical Reaction Balancing
The balanced chemical reaction \(4 \mathrm{Al}(s) + 3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2}\mathrm{O}_{3}(s)\) illustrates the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Balancing a chemical equation ensures that the same number of atoms of each element is present on both sides of the equation.

The coefficients in a balanced equation, like the 4 for aluminum and 3 for oxygen, also give us the stoichiometric ratios of reactants and products involved. These ratios are key to stoichiometry problems because they allow us to convert between moles of different compounds within the same reaction. This is exactly what was done to find out how much aluminum reacted with the given amount of oxygen.
Converting Moles to Mass
When you have the number of moles of a substance, like aluminum in our exercise, converting it to mass is a simple step. Each element has a unique molar mass, which is the weight of one mole of that element. For aluminum, this value is about 26.98 grams per mole.

To convert moles to mass, you use the formula:
\(\text{mass} = \text{moles} \times \text{molar mass}\).
This formula essentially multiplies the abstract moles, which is a count of particles, by the molar mass, which is the weight per count. It is very much like calculating the total weight by knowing how many bags of rice you have and the weight of each bag. In the solution, calculating the mass of aluminum (\(0.119 \, \text{mol}\) times 26.98 g/mol) resulted in the final mass of the reacted aluminum.

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Most popular questions from this chapter

A mixture of \(1.00 \mathrm{~g} \mathrm{H}_{2}\) and \(1.00 \mathrm{~g} \mathrm{He}\) is placed in a \(1.00-\mathrm{L}\) container at \(27^{\circ} \mathrm{C}\). Calculate the partial pressure of each gas and the total pressure.

The average lung capacity of a human is \(6.0 \mathrm{~L}\). How many moles of air are in your lungs when you are in the following situations? a. At sea level \((T=298 \mathrm{~K}, P=1.00 \mathrm{~atm})\). b. \(10 . \mathrm{m}\) below water \((T=298 \mathrm{~K}, P=1.97 \mathrm{~atm})\). c. At the top of Mount Everest \((T=200 . \mathrm{K}, P=0.296 \mathrm{~atm})\).

Atmospheric scientists often use mixing ratios to express the concentrations of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume): ppmv of \(X=\frac{\text { vol of } X \text { at STP }}{\text { total vol of air at STP }} \times 10^{6}\) On a certain November day, the concentration of carbon monoxide in the air in downtown Denver, Colorado, reached \(3.0 \times 10^{2}\) ppmv. The atmospheric pressure at that time was 628 torr and the temperature was \(0^{\circ} \mathrm{C}\). a. What was the partial pressure of \(\mathrm{CO}\) ? b. What was the concentration of \(\mathrm{CO}\) in molecules per cubic meter? c. What was the concentration of \(\mathrm{CO}\) in molecules per cubic centimeter?

A 20.0-L stainless steel container at \(25^{\circ} \mathrm{C}\) was charged with \(2.00\) atm of hydrogen gas and \(3.00\) atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at \(25^{\circ} \mathrm{C} ?\) If the exact same experiment were performed, but the temperature was \(125^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C}\), what would be the pressure in the tank?

A piece of solid carbon dioxide, with a mass of \(7.8 \mathrm{~g}\), is placed in a 4.0-L otherwise empty container at \(27^{\circ} \mathrm{C}\). What is the pressure in the container after all the carbon dioxide vaporizes? If \(7.8 \mathrm{~g}\) solid carbon dioxide were placed in the same container but it already contained air at 740 torr, what would be the partial pressure of carbon dioxide and the total pressure in the container after the carbon dioxide vaporizes?

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