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Chapter 8: Problem 65

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 can be calculated using the following steps: 1. Find the number of moles of oxygen gas: \( \frac{2.00 \, \text{L}}{22.4 \, \text{L/mol}}\) 2. Find the number of moles of aluminum reacted: \( (\text{Moles of O}_2) \times \frac{4 \, \text{moles Al}}{3 \, \text{moles O}_2}\) 3. Calculate the mass of aluminum reacted: \( (\text{Moles of Al}) \times (26.98 \, \text{g/mol})\)

Step by step solution

01

Find the number of moles of oxygen gas

We are given that the volume of oxygen gas at STP is 2.00 L. To find the number of moles, we will use the ideal gas law, which at STP can be simplified to: n = PV/RT At STP, P = 1 atm, V = 2.00 L, R = 0.08206 L atm/mol K, and T = 273.15 K. However, we can simplify the problem since we know 1 mole of any gas occupies 22.4 L at STP. Therefore: Number of moles of oxygen gas = (2.00 L) / (22.4 L/mol)
02

Find the number of moles of aluminum reacted

Now that we have the number of moles of oxygen gas, we can use the stoichiometry of the balanced chemical equation to find the number of moles of aluminum reacted. From the balanced equation, we see that 4 moles of Al reacts with 3 moles of O₂. Hence, we can set up a proportion to determine the moles of aluminum: Moles of Al = (Moles of O₂) × (4 moles Al / 3 moles O₂)
03

Calculate the mass of aluminum reacted

Now that we have the number of moles of aluminum reacted, we can calculate the mass using the molar mass of aluminum, which is approximately 26.98 g/mol. Mass of Al = (Moles of Al) × (26.98 g/mol)

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Most popular questions from this chapter

A \(20.0\) -\(\mathrm{L}\) 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?

Concentrated hydrogen peroxide solutions are explosively decomposed by traces of transition metal ions (such as Mn or Fe): $$2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(i)+\mathrm{O}_{2}(g)$$ What volume of pure \(\mathbf{O}_{2}(g),\) collected at \(27^{\circ} \mathrm{C}\) and 746 torr, would be generated by decomposition of \(125 \mathrm{g}\) of a \(50.0 \%\) by mass hydrogen peroxide solution? Ignore any water vapor that may be present.

An \(11.1\) - \(\mathrm{L}\) sample? sample of gas is determined to contain \(0.50 \) \(\mathrm{mole}\) sample?of \(\mathrm{N}_{2} .\) At the same temperature and pressure, how many moles of gas would there be in a \(20 .\) - \(\mathrm{L}\) sample?

Calculate \(w\) and \(\Delta E\) when \(1\) mole of a liquid is vaporized at its boiling point \(\left(80 .^{\circ} \mathrm{C}\right)\) and \(1.00\) atm pressure. \(\Delta H\) for the vaporization of the liquid is \(30.7 \mathrm{kJ} / \mathrm{mol}\) at \(80 .^{\circ} \mathrm{C}\). Assume the volume of \(1\) mole of liquid is negligible as compared to the volume of \(1\) mole of gas at \(80 .^{\circ} \mathrm{C}\) and \(1.00\) atm.

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\). This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus 1894 p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows:If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.
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