Question

If 10.6 kg of Al2O3(s),10.6 kg of Al2O3(s), 51.4 kg of NaOH(l),51.4 kg of NaOH(l), and 51.4 kg of HF(g)51.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced?

Answer 1

43.668 kg

Step-by-step explanation:

First we set the equation:

`Al_(2)O_(3) + 6NaOH + 12HF \longrightarrow 2Na_(3)AlF_(6)+9H_(2)O`

Now, we need to now the kmoles for each reactant:

`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`

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:

`M_{Na_(3)AlF_(6)}=209.94kg/kmol\\n_{Na_(3)AlF_(6)}=0.208kmol\\m_{Na_(3)AlF_(6)}=43.668kg`