Respuesta :
Answer: The mass of cryolite produced is 51.48 kg
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 aluminium oxide:
Given mass of aluminium oxide = 12.5 kg = 12500 g (Conversion factor: 1 kg = 1000 g)
Molar mass of aluminium oxide = 101.96 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol[/tex]
- For NaOH:
Given mass of NaOH = 55.4 kg = 55400 g
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol[/tex]
- For HF:
Given mass of HF = 55.4 kg = 55400 g
Molar mass of HF = 20 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of HF}=\frac{55400g}{20g/mol}=2770mol[/tex]
For the given chemical reaction:
[tex]Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)[/tex]
By Stoichiometry of the reaction:
1 mole of aluminium oxide reacts with 6 moles of sodium hydroxide and 12 moles of HF.
So, 122.6 moles of aluminium oxide will react with [tex](6\times 122.6)=735.6mol[/tex] of sodium hydroxide and [tex](12\times 122.6)=1471.2mol[/tex] of HF
As, given amount of NaOH and HF is more than the required amount. So, they are considered as an excess reagent.
Thus, aluminium oxide is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of aluminium oxide produces 2 moles of cryolite
So, 122.6 moles of aluminium oxide will produce = [tex]\frac{2}{1}\times 122.6=245.2mol[/tex] of cryolite
Now, calculating the mass of cryolite by using equation 1:
Molar mass of cryolite = 209.94 g/mol
Moles of cryolite = 245.2 mol
Putting values in equation 1, we get:
[tex]245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g[/tex]
Converting this into kilograms, we use the conversion factor:
1 kg = 1000 g
So, [tex]51477.3 g\times (\frac{1kg}{1000g})=51.48kg[/tex]
Hence, the mass of cryolite produced is 51.48 kg
The mass of cryolite that would be produced is 51.5 kg
From the question,
We are to determine the mass of cryolite that would be produced
First,
We will write the balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
Al₂O₃(s) + 6NaOH(l) + 12HF(g) → 2Na₃AlF₆ + 9H₂O(g)
This means
1 mole of Al₂O₃ reacts with 6 moles of NaOH and 12 moles of HF to produce 2 moles of Na₃AlF₆ and 9 moles H₂O
To determine the mass of cryolite (Na₃AlF₆) that would be produced,
First, we will determine the number of moles of each reactant present
- For aluminum oxide (Al₂O₃)
Mass = 12.5 kg = 12500 g
Molar mass = 101.96 g/mol
From the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
∴ Number of moles of Al₂O₃ present = [tex]\frac{12500}{101.96}[/tex]
Number of moles of Al₂O₃ present = 122.597 moles
- For NaOH
Mass = 55.4 kg = 55400 g
Molar mass = 39.997 g/mol
Number of moles of NaOH = [tex]\frac{55400}{39.997}[/tex]
∴ Number of moles of NaOH present = 1385.10 moles
- For HF
Mass = 55.4 kg = 55400 g
Molar mass = 20.01 g/mol
Number of moles of HF = [tex]\frac{55400}{20.01 }[/tex]
∴ Number of moles of HF present = 2768.61 moles
Now,
Since
1 mole of Al₂O₃ reacts with 6 moles of NaOH and 12 moles of HF to produce 2 moles of Na₃AlF₆
Then,
122.597 moles of Al₂O₃ reacts with 735.582 moles of NaOH and 1471.164 moles of HF to produce 245.194 moles of Na₃AlF₆
∴ The number of moles of cryolite produced is 245.194 moles
Now, for the mass of cryolite that would be produced,
From the formula
Mass = Number of moles × Molar mass
Molar mass of cryolite (Na₃AlF₆) = 209.94 g/mol
∴ Mass of cryolite that would be produced = 245.194 × 209.94
Mass of cryolite that would be produced = 51476.028 g
Mass of cryolite that would be produced = 51.476028 kg
Mass of cryolite that would be produced ≅ 51.5 kg
Hence, the mass of cryolite that would be produced is 51.5 kg
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