Question

If 13.6 kilograms of al2o3 51.4 kilograms of naoh and 51.4 kilograms of hf react completely, how many kilograms of cryolite will be produced

Answer 1

The chemical reaction that occurs between $Al_2O_3$, $NaOH$ and $HF$ to form cryolite (molecular formula: $Na_3AlF_6$) is:

$Al_2O_3+NaOH+HF\rightarrow Na_3AlF_6+H_2O$

The balanced reaction is:

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

In order to determine the weight of cryolite produced, the limiting reactant (the reactant that is completely used up in the reaction) should be identified first. To determine the limiting reactant, the number of moles of each reactant with respect to cryolite ($2Na_3AlF_6$) should be calculated.

The formula for determining the number of moles is:

$number of mole = \frac{given mass}{molar mass}$

1. For $Al_2O_3$:

According to the balanced equation, 1 mole of $Al_2O_3$ reacts to give 2 moles of $Na_3AlF_6$.

Molar mass of $Al_2O_3 = 101.96 g/mol$

So, $\frac{13.6 \times 10^{3} g Al_2O_3}{101.96 g/mol}\times 2 moles of Na_3AlF_6$

Number of moles = $2.668\times 10^{2} mole$  $Na_3AlF_6$

2. For $NaOH$ :

According to the balanced equation, 6 mole of $NaOH$ reacts to give 2 moles of $Na_3AlF_6$.

Molar mass of $NaOH = 39.997 g/mol$

So, $\frac{51.4 \times 10^{3} g NaOH}{39.997 g/mol}\times \frac{2 moles of Na_3AlF_6}{6 moles of NaOH}$

The number of moles = $4.279\times 10^{2} mole$  $Na_3AlF_6$

3. For $HF$ :

According to the balanced equation, 12 moles of $HF$ reacts to give 2 moles of $Na_3AlF_6$.

Molar mass of $HF = 12.01 g/mol$

So, $\frac{51.4 \times 10^{3} g HF}{20.01 g/mol}\times \frac{2 moles of Na_3AlF_6}{12 moles of HF}$

Number of moles = $4.282\times 10^{2} mole$ $Na_3AlF_6$.

Thus, $Al_2O_3$ is the limiting reactant.

Molar mass of $Na_3AlF_6 = 209.94 g/mol$

Amount of $Na_3AlF_6$ produced = $2.668\times 10^{2} mole\times 209.94 g/mol$ = $5.6012\times 10^{4} g$

Since, $1 kg = 1000 g$

$5.6012\times 10^{4} g = 56.012 kg$

Hence, the amount of cryolite produced is $56.012 kg$.