Question

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation for the synthesis of cryolite. equation: Al2O3 (s) + NaOH (l) + HF (g) -> Na3AlF6 + H2O (g) Al2O3(s)+NaOH(l)+HF(g)⟶Na3AlF6+H2O(g) If 13.4 kg of Al2O3(s), 55.4 kg of NaOH(l), and 55.4 kg of HF(g) react completely, how many kilograms of cryolite will be produced?

Answer 1

`55.2kgNa_(3)AlF_(6)`

Step-by-step explanation:

1. First balance the equation for the synthesis of cryolite:

`Al_(2)O_(3)_((s))+6NaOH_((l))+12HF_((g))=2Na_(3)AlF_(6)+9H_(2)O_((g))`

2. Find the limiting reagent between the
`Al_(2)O_(3),NaOH` and
`HF`

- First calculate the number of moles of each compound using its molar mass and the mass that reacted completely:

`13.4kgAl_(2)O_(3)*(1molAl_(2)O_(3))/(101.96gAl_(2)O_(3))*(1000g)/(1kg)=131molesAl_(2)O_(3)`

`55.4kgNaOH*(1molNaOH)/(40kgNaOH)*(1000g)/(1kg)=1385molesNaOH`
`55.4kgHF*(1molHF)/(20kgHF)*(1000g)/(1kg)=2770molesHF`

- Divide the number of moles obtained between the stoichiometric coefficient of each compound in the chemical reaction:

`Al_(2)O_(3):(131)/(1)=131`

`NaOH:(1385)/(6)=231`

`HF:(2770)/(12)=231`

The
`Al_(2)O_(3)` is the limiting reagent because it has the smallest number.

3. Find the mass of cryolite produced:

`13.4kgAl_(2)O_(3)*(1molAl_(2)O_(3))/(0.10196kgAl_(2)O_(3))*(2molesNa_(3)AlF_(6))/(1molAl_(2)O_(3))*(0.20994kgNa_(3)AlF_(6))/(1molNa_(3)AlF_(6))=55.2kgNa_(3)AlF_(6)`