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 $HF$.

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

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

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

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 $129.54\times 2=259.08$ moles of $Na_3AlF_6$

Now we have to calculate the mass of $Na_3AlF_6$

$\text{ Mass of }Na_3AlF_6=\text{ Moles of }Na_3AlF_6\times \text{ Molar mass of }Na_3AlF_6$

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

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