Cryolite, na3alf6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. balance the equation for 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?

The mass of cryolite produced is 51.48 kg

Explanation:

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

     .....(1)

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:

Given mass of NaOH = 55.4 kg = 55400 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

Given mass of HF = 55.4 kg = 55400 g

Molar mass of HF = 20 g/mol

Putting values in equation 1, we get:

For the given chemical reaction:

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 of sodium hydroxide and 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 = 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:

Converting this into kilograms, we use the conversion factor:

1 kg = 1000 g

So,

Hence, the mass of cryolite produced is 51.48 kg

c. 75.7

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