Respuesta :
Answer:
For A: The mass of aluminium required will be 183 g
For B: The mass of alumina produced will be [tex]6.63\times 10^6g[/tex]
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For a:
Given mass of [tex]NH_4ClO_4[/tex] = 1.325 kg = 1325 g (Conversion factor: 1 kg = 1000 g)
Molar mass of [tex]NH_4ClO_4[/tex] = 117.50 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }NH_4ClO_4=\frac{1325g}{117.50g/mol}=11.28mol[/tex]
For the given chemical reaction:
[tex]6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)[/tex]
By stoichiometry of the reaction:
6 moles of [tex]NH_4ClO_4[/tex] reacts with 10 moles of aluminium
So, 11.28 moles of [tex]NH_4ClO_4[/tex] will react with = [tex]\frac{10}{6}\times 11.28=6.77mol[/tex] 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:
[tex]6.77mol=\frac{\text{Mass of aluminium}}{27.00g/mol}\\\\\text{Mass of aluminium}=(6.77mol\times 27.00g/mol)=183g[/tex]
Hence, the mass of aluminium required will be 183 g
- For b:
Given mass of aluminium = [tex]3.500\times 10^3=3.5\times 10^6g[/tex] (Conversion factor: 1 kg = 1000 g)
Molar mass of aluminium = 27.00 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of aluminium}=\frac{3.5\times 10^6g}{27.00g/mol}=1.3\times 10^5mol[/tex]
For the given chemical reaction:
[tex]6NH_4ClO_4(s)+10Al(s)\rightarrow 5Al_2O_3(s)+3N_2(g)+6HCl(g)+9H_2O(g)[/tex]
By stoichiometry of the reaction:
10 moles of aluminium produces 5 moles of alumina
So, [tex]1.3\times 10^5mol[/tex] of aluminium will react with = [tex]\frac{5}{10}\times 1.3\times 10^5=6.5\times 10^4mol[/tex] 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 = [tex]6.5\times 10^4mol[/tex]
Putting values in equation 1, we get:
[tex]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[/tex]
Hence, the mass of alumina produced will be [tex]6.63\times 10^6g[/tex]
183g mass of aluminum should be mixed with 1.325 kg of NH₄ClO₄ and 6.63 10⁶ g of alumina produced.
How we calculate moles?
Moles of any substance can be calculated as:
n = W/M, where
W = given mass
M = molar mass
Given chemical reaction is:
6NH₄ClO₄(s) + 10Al(s) → 5Al₂O₃(s) + 3N₂(g) + 6HCl(g) + 9H₂O(g)
From the stoichiometry of the reaction:
6 moles of NH₄ClO₄ = react with 10 moles of Aluminum
1 mole of NH₄ClO₄ = react with 10/6 moles of Aluminum
Given mass of NH₄ClO₄ = 1.325 kg
Required moles of NH₄ClO₄ can be calculated by using the above formula = 1.325 / 117.50 = 11.28 mole
11.28 mole of NH₄ClO₄ = react with 10/6 × 11.28 = 6.77 moles of Aluminum
Mass of aluminum can be calculated by using the above formula:
Mass of aluminum = 6.77 × 27 = 183 g.
From the stoichiometry of the reaction:
10 moles of Al = react with 5 moles of Al₂O₃
1 mole of Al = react with 5/10 moles of Al₂O₃
Given mass of Al = 3.500 × 10³ kg = 3.5 × 10⁶ g
Required moles of Al can be calculated by using the above formula = 3.5×10⁶/ 27 = 1.3 × 10⁵ mole
1.3 × 10⁵ mole of Al = react with 5/10 × 1.3×10⁵= 6.5×10⁴ moles of alumina
Mass of alumina can be calculated by using the above formula:
Mass of alumina = 6.5×10⁴ × 101.96 = 6.63 10⁶ g.
Hence, required mass of aluminum is 183 g and mass of alumina is 6.63 10⁶ g.
To know more about moles, visit the below link:
https://brainly.com/question/24639749
