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Chapter 5: Problem 71

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 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 that reacted with 2.00 L of pure oxygen gas at STP is 3.214 g.

Step by step solution

01

Determine the number of moles of oxygen gas

Using the volume of oxygen gas and the conditions at STP (Standard Temperature and Pressure), we can find the number of moles of oxygen gas. Recall that at STP, 1 mole of any gas occupies a volume of 22.4 L. Thus, we can find the moles of oxygen gas using the following formula: Moles of O₂ = Volume of O₂ / Volume occupied by 1 mole of O₂ at STP Moles of O₂ = 2.00 L / 22.4 L/mol Moles of O₂ = 0.0893 mol
02

Determine the number of moles of aluminum

Now that we know the number of moles of oxygen gas, we can use the stoichiometry of the balanced equation to find the number of moles of aluminum required for the reaction: 4 Al + 3 O₂ → 2 Al₂O₃ From the balanced equation, we can see that 3 moles of O₂ react with 4 moles of Al. Using this ratio, we can calculate the number of moles of aluminum: Moles of Al = (moles of O₂ × moles of Al) / moles of O₂ Moles of Al = (0.0893 mol × 4 mol) / 3 mol Moles of Al = 0.1191 mol
03

Convert the number of moles of aluminum to mass

Now that we have found the number of moles of aluminum, we can convert it to mass using the molar mass of aluminum (26.98 g/mol): Mass of Al = moles of Al × molar mass of Al Mass of Al = 0.1191 mol × 26.98 g/mol Mass of Al = 3.214 g Therefore, the mass of aluminum that reacted with oxygen gas is 3.214 g.

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Most popular questions from this chapter

Silane, SiH, , is the silicon analogue of methane, \(\mathrm{CH}_{4}\) . It is prepared industrially according to the following equations: $$\begin{array}{c}{\mathrm{Si}(s)+3 \mathrm{HCl}(g) \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g)} \\ {4 \mathrm{HSiCl}_{3}(l) \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l)}\end{array}$$ a. If \(156 \mathrm{mL}\) \(\mathrm{HSiCl}_{3} (d=1.34 \mathrm{g} / \mathrm{mL})\) is isolated when 15.0 \(\mathrm{L}\) \(\mathrm{HCl}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3} ?\) b. When \(156 \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is 93.1\(\% ?\)

Do all the molecules in a 1 -mole sample of \(\mathrm{CH}_{4}(g)\) have the same kinetic energy at 273 \(\mathrm{K}\)? Do all molecules in a 1 -mole sample of \(\mathrm{N}_{2}(g)\) have the same velocity at 546 \(\mathrm{K}\) ? Explain.

At \(0^{\circ} \mathrm{C}\) a \(1.0-\mathrm{L}\) flask contains $5.0 \times 10^{-2}\( mole of \)\mathrm{N}_{2}, 1.5 \times\( \)10^{2} \mathrm{mg} \mathrm{O}_{2},\( and \)5.0 \times 10^{21}\( molecules of \)\mathrm{NH}_{3} .$ What is the partial pressure of each gas, and what is the total pressure in the flask?

Methanol (CH_l3 \(\mathrm{OH}\) ) can be produced by the following reaction: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$ Hydrogen at STP flows into a reactor at a rate of 16.0 L/min. Carbon monoxide at STP flows into the reactor at a rate of 25.0 L/min. If 5.30 g methanol is produced per minute, what is the percent yield of the reaction?

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?
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