To determine the mass of ammonium perchlorate ([tex]\(NH_4ClO_4\)[/tex]) required for every kilogram of aluminum (Al) in the fuel mixture, follow this step-by-step process:
1. Determine the molar masses of aluminum and ammonium perchlorate:
- Molar mass of Al (Aluminum): 26.98 g/mol
- Molar mass of [tex]\(NH_4ClO_4\)[/tex] (Ammonium Perchlorate): 117.49 g/mol
2. Convert the given mass of aluminum to moles:
- Given mass of Al = 1000 grams (1 kilogram)
- Moles of Al = mass of Al / molar mass of Al
[tex]\[
\text{Moles of Al} = \frac{1000 \text{ g}}{26.98 \text{ g/mol}} \approx 37.064 \text{ moles}
\][/tex]
3. Determine the molar ratio from the balanced chemical equation:
From the balanced equation:
[tex]\[
3 \text{ Al} + 3 \text{ NH}_4\text{ClO}_4 \longrightarrow \text{Products}
\][/tex]
The molar ratio of Al to [tex]\(NH_4ClO_4\)[/tex] is 1:1.
4. Calculate the moles of [tex]\(NH_4ClO_4\)[/tex] required:
Given the 1:1 molar ratio:
[tex]\[
\text{Moles of } NH_4ClO_4 = \text{Moles of Al} \approx 37.064 \text{ moles}
\][/tex]
5. Convert moles of [tex]\(NH_4ClO_4\)[/tex] to mass:
- Mass of [tex]\(NH_4ClO_4\)[/tex] = moles of [tex]\(NH_4ClO_4\)[/tex] [tex]\(\times\)[/tex] molar mass of [tex]\(NH_4ClO_4\)[/tex]
[tex]\[
\text{Mass of } NH_4ClO_4 = 37.064 \text{ moles} \times 117.49 \text{ g/mol} \approx 4354.707 \text{ grams}
\][/tex]
So, for every kilogram of aluminum, you should use approximately 4354.707 grams (or approximately 4.354 kilograms) of ammonium perchlorate in the fuel mixture.
