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

Step-by-step 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}=(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 =
`(2)/(1)* 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* 210g/mol)=1181.8g`

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