Question

If 13.2 kg of Al 2 O 3 ( s ) , 50.4 kg of NaOH ( l ) , and 50.4 kg of HF ( g ) react completely, how many kilograms of cryolite will be produced? mass of cryolite produced: kg Na 3 AlF 6

Answer 1

Answer : The mass of cryolite produced will be, 54.38 kg

Solution : Given,

Mass of
`Al_2O_3` = 13.2 kg = 13200 g

Mass of
`NaOH` = 50.4 kg = 50400 g

Mass of
`HF` = 50.4 kg = 50400 g

Molar mass of
`Al_2O_3` = 101.9 g/mole

Molar mass of
`NaOH` = 40 g/mole

Molar mass of
`HF` = 20 g/mole

Molar mass of
`Na_3AlF_6` = 209.9 g/mole

First we have to calculate the moles of
`Al_2O_3,NaOH` and
`HF`.

`\text{ Moles of }Al_2O_3=\frac{\text{ Mass of }Al_2O_3}{\text{ Molar mass of }Al_2O_3}=(13200g)/(101.9g/mole)=129.54moles`

`\text{ Moles of }NaOH=\frac{\text{ Mass of }NaOH}{\text{ Molar mass of }NaOH}=(50400g)/(40g/mole)=1260moles`

`\text{ Moles of }HF=\frac{\text{ Mass of }HF}{\text{ Molar mass of }HF}=(50400g)/(20g/mole)=2520moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

The mole ratio of
`Al_2O_3,NaOH` and
`HF` is, 1 : 6 : 12

And, the ratio of given moles of
`Al_2O_3,NaOH` and
`HF` is, 129.54 : 1260 : 2520

From this we conclude that,
`NaOH,HF` is an excess reagent because the given moles are greater than the required moles and
`Al_2O_3` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`Na_3AlF_6`

From the reaction, we conclude that

As, 1 mole of
`Al_2O_3` react to give 2 mole of
`Na_3AlF_6`

So, 129.54 moles of
`Al_2O_3` react to give
`129.54* 2=259.08` moles of
`Na_3AlF_6`

Now we have to calculate the mass of
`Na_3AlF_6`

`\text{ Mass of }Na_3AlF_6=\text{ Moles of }Na_3AlF_6* \text{ Molar mass of }Na_3AlF_6`

`\text{ Mass of }Na_3AlF_6=(259.08moles)* (209.9g/mole)=54380.892g=54.38kg`

Thus, the mass of cryolite produced will be, 54.38 kg