Question

The solid fuel in the booster stage of the space shuttle is a mix- ture of ammonium perchlorate and aluminum powder. Upon igni- tion, the reaction that takes place is 6 NH4ClO4(s) 1 10 Al(s) S 5 Al2O3(s) 1 3 N2(g) 1 6 HCl(g) 1 9 H2O(g). (a) What mass of aluminum should be mixed with 1.325 kg of NH4ClO4 for this reaction? (b) Determine the mass of Al2O3 (alumina, a finely divided white powder that is produced as billows of white smoke) formed in the reaction of 3.500 3 103 kg of aluminum.

Answer 1

Answer:

For A: The mass of aluminium required will be 183 g

For B: The mass of alumina produced will be $6.63\times 10^6g$

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For a:

Given mass of $NH_4ClO_4$ = 1.325 kg = 1325 g   (Conversion factor:  1 kg = 1000 g)

Molar mass of $NH_4ClO_4$ = 117.50 g/mol

Putting values in equation 1, we get:

$\text{Moles of }NH_4ClO_4=\frac{1325g}{117.50g/mol}=11.28mol$

For the given chemical reaction:

$6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)$

By stoichiometry of the reaction:

6 moles of $NH_4ClO_4$ reacts with 10 moles of aluminium

So, 11.28 moles of $NH_4ClO_4$ will react with = $\frac{10}{6}\times 11.28=6.77mol$ of aluminium

Now, calculating the mass of aluminium by using equation 1, we get:

Molar mass of aluminium = 27.00 g/mol

Moles of aluminium = 6.77 moles

Putting values in equation 1, we get:

$6.77mol=\frac{\text{Mass of aluminium}}{27.00g/mol}\\\\\text{Mass of aluminium}=(6.77mol\times 27.00g/mol)=183g$

Hence, the mass of aluminium required will be 183 g

• For b:

Given mass of aluminium = $3.500\times 10^3=3.5\times 10^6g$   (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium = 27.00 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium}=\frac{3.5\times 10^6g}{27.00g/mol}=1.3\times 10^5mol$

For the given chemical reaction:

$6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)$

By stoichiometry of the reaction:

10 moles of aluminium produces 5 moles of alumina

So, $1.3\times 10^5mol$ of aluminium will react with = $\frac{5}{10}\times 1.3\times 10^5=6.5\times 10^4mol$ of alumina

Now, calculating the mass of alumina by using equation 1, we get:

Molar mass of alumina = 101.96 g/mol

Moles of alumina = $6.5\times 10^4mol$

Putting values in equation 1, we get:

$6.5\times 10^4mol=\frac{\text{Mass of alumina}}{101.96g/mol}\\\\\text{Mass of alumina}=(6.5\times 10^4mol\times 101.96g/mol)=6.63\times 10^6g$

Hence, the mass of alumina produced will be $6.63\times 10^6g$