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Chapter 3: Problem 111

The reusable booster rockets of the U.S. space shuttle employ a mixture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is $$ 3 \mathrm{Al}(s)+3 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) \longrightarrow $$ $$ \mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{AlCl}_{3}(s)+3 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ What mass of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) should be used in the fuel mixture for every kilogram of Al?

Short Answer

Expert verified
To determine the mass of ammonium perchlorate (NH₄ClO₄) needed for every kilogram of aluminum (Al), we calculate the mass ratio between Al and NH₄ClO₄ using their molar masses and the balanced chemical equation. We find that for 1 kg of Al, approximately 4.36 kg of NH₄ClO₄ is needed in the fuel mixture.

Step by step solution

01

Identify the balanced chemical equation

The balanced chemical equation for the reaction between aluminum (Al) and ammonium perchlorate (NH₄ClO₄) is given by: \[ 3 \mathrm{Al}(s)+3 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{AlCl}_{3}(s)+3 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \]
02

Calculate the mole ratio between Al and NH₄ClO₄

From the balanced chemical equation, we can see that the mole ratio between Al and NH₄ClO₄ is 3:3, which can be simplified to 1:1. This means that for every 1 mole of aluminum, 1 mole of ammonium perchlorate is required.
03

Find the molar masses of Al and NH₄ClO₄

We need to determine the molar masses of Al and NH₄ClO₄ to convert the mole ratio into a mass ratio. - The molar mass of Al (Aluminum) is 26.98 g/mol. - The molar mass of NH₄ClO₄ (Ammonium perchlorate) can be calculated as follows: - N (Nitrogen) has a molar mass of 14.01 g/mol. - H (Hydrogen) has a molar mass of 1.008 g/mol. - Cl (Chlorine) has a molar mass of 35.45 g/mol. - O (Oxygen) has a molar mass of 16.00 g/mol. NH₄ClO₄ = (1 × 14.01) + (4 × 1.008) + (1 × 35.45) + (4 × 16.00) = 117.53 g/mol
04

Calculate the mass ratio between Al and NH₄ClO₄

We know that the mole ratio between Al and NH₄ClO₄ is 1:1. Now, we can calculate the mass ratio using the molar masses of Al and NH₄ClO₄: For every 1 mole of Al (26.98 g), we need 1 mole of NH₄ClO₄ (117.53 g). So, the mass ratio between aluminum and ammonium perchlorate is 26.98:117.53.
05

Determine the mass of NH₄ClO₄ needed for 1 kg of Al

We need to calculate the mass of NH₄ClO₄ required for every kilogram (1000 grams) of Al. We can use the mass ratio of Al and NH₄ClO₄ to determine this: For 26.98 g of Al, we need 117.53 g of NH₄ClO₄. For 1 g of Al, we need \(\frac{117.53}{26.98} \) g of NH₄ClO₄. For 1000 g (1 kg) of Al, we need \(\frac{117.53}{26.98} \times 1000 \)g of NH₄ClO₄. Thus, the required mass of NH₄ClO₄ for 1 kg of Al is approximately 4355.41 g or 4.36 kg.

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Most popular questions from this chapter

Aluminum reacts with bromine, producing aluminum bromide: $$ 2 \mathrm{Al}(s)+3 \mathrm{Br}_{2}(l) \rightarrow 2 \mathrm{AlBr}_{3}(s) $$ In a certain experiment, 20.0 \(\mathrm{mL}\) of bromine (density \(=\) 3.10 \(\mathrm{g} / \mathrm{mL}\) ) was reacted with excess aluminum to yield 50.3 \(\mathrm{g}\) of aluminum bromide. What is the percent yield for this experiment?

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g)\) according to these unbalanced equations: $$ \begin{array}{l}{\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)} \\\ {\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)}\end{array} $$ In a certain experiment 2.00 moles of \(\mathrm{NH}_{3}(g)\) and 10.00 moles of \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)

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The compound cisplatin, $\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2},$ has been studied as an antitumor agent. Cisplatin is synthesized as follows: $$ \mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \rightarrow \operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q) $$ What mass of cisplatin can be produced from \(100 .\) g of $\mathrm{K}_{2} \mathrm{PtCl}_{4}\( and sufficient \)\mathrm{NH}_{3} ?$

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