Answer:
43.668 kg
Explanation:
First we set the equation:
[tex]Al_{2}O_{3} + 6NaOH + 12HF \longrightarrow 2Na_{3}AlF_{6}+9H_{2}O[/tex]
Now, we need to now the kmoles for each reactant:
[tex]M_{Al_{2}O_{3}}=101.96kg/kmol\\M_{NaOH}=40kg/kmol\\M_{HF}=20.01kg/kmol\\n_{Al_{2}O_{3}}=0.104kmol\\n_{NaOH}=1.285kmol\\n_{HF}=2.57kmol[/tex]
With this, we can see that the limit reactant is the aluminum oxide, so, with the equation for the reaction we know that 1 kmol of aluminum oxide, produces 2 kmol of cryolite, so we set a rule of three and see that 0.208 kmoles of cryolite are produced, the we proceed to calculate the mass:
[tex]M_{Na_{3}AlF_{6}}=209.94kg/kmol\\n_{Na_{3}AlF_{6}}=0.208kmol\\m_{Na_{3}AlF_{6}}=43.668kg[/tex]
