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Chapter 9: Problem 110

Aluminum metal burns in chlorine gas to form aluminum chloride. (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting if \(100.0 \mathrm{~g}\) of \(\mathrm{Al}\) and \(5.00\) moles of \(\mathrm{Cl}_{2}\) are used?

Short Answer

Expert verified
(a) The balanced chemical equation for the reaction is: \[ 2 \mathrm{Al} \ + \ 3 \mathrm{Cl_{2}} \ \longrightarrow \ 6 \mathrm{AlCl_{3}} \] (b) Cl2 is the limiting reactant.

Step by step solution

01

Write the unbalanced chemical equation

First, let's write down the unbalanced chemical equation for the reaction between aluminum (Al) and chlorine gas (Cl2) to form aluminum chloride (AlCl3): \[ \mathrm{Al} \ + \ \mathrm{Cl_{2}} \ \longrightarrow \ \mathrm{AlCl_{3}} \]
02

Balance the chemical equation

Now, we need to balance the chemical equation. We have one Al on both sides, but we have two Cl atoms on the left side and three on the right side. To balance the Cl atoms, we can place a 3 in front of Cl2 and a 2 in front of AlCl3: \[ 2 \mathrm{Al} \ + \ 3 \mathrm{Cl_{2}} \ \longrightarrow \ 6 \mathrm{AlCl_{3}} \]
03

Calculate the moles of each reactant

Before determining the limiting reactant, we need to find out how many moles of each reactant we have. For Al, we are given 100.0 g. We can use the molar mass of Al to convert grams to moles: \[ \mathrm{moles\ of\ Al} = \frac{100.0 \mathrm{~g}}{26.98 \mathrm{~g/mol}} \] For Cl2, we are given 5.00 moles directly.
04

Calculate the moles needed for the reaction

From the balanced chemical equation, for every 2 moles of Al, 3 moles of Cl2 are required. Let x and y represent the moles of Al and Cl2 used in the reaction, respectively. \(\frac{x}{2} = \frac{y}{3}\) Now substitute the given amounts and solve. \(\frac{x}{2\mathrm{(moles)}} = \frac{5.00\mathrm{(moles)}}{3}\)
05

Determine the limiting reactant

Solving for x gives: \[ x = \frac{10.00}{3} \mathrm{~moles} \] To react with all the given Cl2, we need only 10/3 moles of Al. But we have 100.0 g of Al which is more than 10/3 moles. Therefore, Cl2 is the limiting reactant, as it runs out first.

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Most popular questions from this chapter

Consider the unbalanced chemical equation \(\mathrm{KCl}+\mathrm{O}_{2} \rightarrow \mathrm{KClO}_{3}\) If you react \(42.6 \mathrm{~g}\) of \(\mathrm{KCl}\) and \(36.5 \mathrm{~g}\) of \(\mathrm{O}_{2}\) and the reaction has a percent yield of \(56.0 \%\), how many grams of \(\mathrm{KClO}_{3}\) are produced?

Determine the empirical formula of the compound that is \(43.2 \%\) by mass \(\mathrm{K}, 39.1 \%\) by \(\mathrm{mass} \mathrm{Cl}\), and also contains oxygen.

Lead(II) sulfide reacts with hydrogen peroxide to give lead(II) sulfate as shown in the unbalanced chemical equation \(\mathrm{PbS}+\mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{PbSO}_{4}+\mathrm{H}_{2} \mathrm{O}\) If \(63.2 \mathrm{~g}\) of \(\mathrm{PbS}\) is reacted with \(48.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}_{2}\) (a) Which is the limiting reagent? (b) How many grams of excess reactant remain after the reaction is complete?

Consider the balanced chemical equation \(\mathrm{SCl}_{4}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{SO}_{2}+4 \mathrm{HCl}\) (a) How many grams of \(\mathrm{H}_{2} \mathrm{O}\) will react with \(5.000 \mathrm{~g}\) of \(\mathrm{SCl}_{4} ?\) (b) How many grams of \(\mathrm{SO}_{2}\) can you make from \(10.00 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} ?\) (c) Suppose you react \(5.000 \mathrm{~g}\) of \(\mathrm{SCl}_{4}\) with the amount of \(\mathrm{H}_{2} \mathrm{O}\) you calculated in part (a). How many grams of \(\mathrm{HCl}\) will you form?

Vitamin \(\mathrm{C}\), also known as ascorbic acid, contains carbon, hydrogen, and possibly oxygen. A \(0.160 \mathrm{~g}\) sample of ascorbic acid is subjected to combustion analysis, yielding \(40.93 \%\) C and \(4.58 \%\) H. If the molar mass of ascorbic acid is approximately \(176 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?
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