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KIM [24]
3 years ago
14

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f

or the synthesis of cryolite. equation: Al2O3 (s) + 6NaOH (l) + 12HF (g) -> 2 Na3AlF6 + 9H2O (g) Al2O3(s)+6NaOH(l)+12HF(g)⟶2Na3AlF6+9H2O(g) If 12.5 kg of Al2O3(s), 55.4 kg of NaOH(l), and 55.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced?
Chemistry
1 answer:
dmitriy555 [2]3 years ago
7 0

<u>Answer:</u> The mass of cryolite produced is 51.48 kg

<u>Explanation:</u>

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

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

  • <u>For aluminium oxide:</u>

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:

\text{Moles of aluminium oxide}=\frac{12500g}{101.96g/mol}=122.6mol

  • <u>For NaOH:</u>

Given mass of NaOH = 55.4 kg = 55400 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of NaOH}=\frac{55400g}{40g/mol}=1389mol

  • <u>For HF:</u>

Given mass of HF = 55.4 kg = 55400 g

Molar mass of HF = 20 g/mol

Putting values in equation 1, we get:

\text{Moles of HF}=\frac{55400g}{20g/mol}=2770mol

For the given chemical reaction:

Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)

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 (6\times 122.6)=735.6mol of sodium hydroxide and (12\times 122.6)=1471.2mol 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 = \frac{2}{1}\times 122.6=245.2mol 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:

245.2mol=\frac{\text{Mass of cryolite}}{209.94g/mol}\\\\\text{Mass of cryolite}=(245.2mol\times 209.94g/mol)=51477.3g

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So, 51477.3 g\times (\frac{1kg}{1000g})=51.48kg

Hence, the mass of cryolite produced is 51.48 kg

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