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Chapter 5: Problem 107

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 $$\begin{aligned}3 \mathrm{Al}(s)+3 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) & \longrightarrow \\ \mathrm{Al}_{2} \mathrm{O}_{3}(s)+& \mathrm{AlCl}_{3}(s)+3 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\end{aligned}$$ 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
For every kilogram of aluminum, 4350.26 g of ammonium perchlorate should be used in the fuel mixture.

Step by step solution

01

1. Write down the balanced chemical equation.

The balanced chemical equation for the reaction of aluminum with ammonium perchlorate is: \(3 \mathrm{Al}(s) + 3 \mathrm{NH}_{4}\mathrm{ClO}_{4}(s) \rightarrow \mathrm{Al}_{2}\mathrm{O}_{3}(s) + \mathrm{AlCl}_{3}(s) + 3 \mathrm{NO}(g) + 6 \mathrm{H}_{2}\mathrm{O}(g) \)
02

2. Calculate the molar mass of aluminum and ammonium perchlorate.

First, we need to calculate the molar masses of aluminum (Al) and ammonium perchlorate (NH4ClO4). - Molar mass of Al: \(26.98 \,g/mol\) - Molar mass of NH4ClO4: \((1 × 14.01) + (4 × 1.01) + (1 × 35.45) + (4 × 16.00) = 117.49 \, g/mol\)
03

3. Use the stoichiometry of the balanced equation to find the mass ratio of NH4ClO4 to Al.

From the balanced equation, we see that 3 moles of Al react with 3 moles of NH4ClO4. Therefore, the mole ratio of NH4ClO4 to Al is 1:1. To find the mass ratio, we'll use the molar masses we calculated in the previous step: Mass ratio: \(\frac{117.49 \, g/mol(NH_{4}ClO_{4})}{26.98 \, g/mol(Al)}= \frac{mass \, of \, NH_{4}ClO_{4}}{mass \, of \, Al} \)
04

4. Calculate the mass of NH4ClO4 needed for every kilogram of Al.

Let's plug the mass of Al (1 kg, or 1000 g) into the mass ratio equation and solve for the mass of NH4ClO4: \(\frac{117.49 \, g/mol(NH_{4}ClO_{4})}{26.98 \, g/mol(Al)}= \frac{mass \, of \, NH_{4}ClO_{4}}{1000 \, g(Al)} \) Now, solve for the mass of NH4ClO4: \(mass \, of \, NH_{4}ClO_{4} = \frac{117.49 \, g/mol(NH_{4}ClO_{4})}{26.98 \, g/mol(Al)} × 1000 \, g(Al)\) \(mass \, of \, NH_{4}ClO_{4} = 4350.26 \, g\) For every kilogram of aluminum, 4350.26 g of ammonium perchlorate should be used in the fuel mixture.

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Most popular questions from this chapter

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One of relatively few reactions that takes place directly between two solids at room temperature is $$\begin{aligned}\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{NH}_{4} \mathrm{SCN}(s) & \longrightarrow \\\\\mathrm{Ba}(\mathrm{SCN})_{2}(s) &+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(g)\end{aligned}$$ In this equation, the \(\cdot 8 \mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) indicates the presence of eight water molecules. This compound is called barium hydroxide octahydrate. a. Balance the equation. b. What mass of ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\) must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?

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