Question

Cryolite (Na3AlF6) is used in the production of aluminum from its ores. It is made by the reaction 6 NaOH 1 Al2O3 1 12 HF 8n 2 Na3AlF6 1 9 H2O Calculate the mass of cryolite that can be prepared by the complete reaction of 287 g Al2O3.

Answer 1

Answer: The mass of cryolite produced in the reaction is 1181.8 g

Explanation:

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

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

Given mass of aluminium oxide = 287 g

Molar mass of aluminium oxide = 102 g/mol

Putting values in equation 1, we get:

$\text{Moles of aluminium oxide}=\frac{287g}{102g/mol}=2.814mol$

The given chemical reaction follows:

$6NaOH+Al_2O_3+12HF\rightarrow 2Na_3AlF_6+9H_2O$

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 2.814 moles of aluminium oxide will produce = $\frac{2}{1}\times 2.814=5.628mol$ of cryolite

Now, calculating the mass of cryolite by using equation 1:

Moles of cryolite = 5.628 moles

Molar mass of cryolite = 210 g/mol

Putting values in equation 1, we get:

$5.628mol=\frac{\text{Mass of cryolite}}{210g/mol}\\\\\text{Mass of cryolite}=(5.628mol\times 210g/mol)=1181.8g$

Hence, the mass of cryolite produced in the reaction is 1181.8 g