Answer:
[tex]55.2kgNa_{3}AlF_{6}[/tex]
Explanation:
1. First balance the equation for the synthesis of cryolite:
[tex]Al_{2}O_{3}_{(s)}+6NaOH_{(l)}+12HF_{(g)}=2Na_{3}AlF_{6}+9H_{2}O_{(g)}[/tex]
2. Find the limiting reagent between the [tex]Al_{2}O_{3},NaOH[/tex] and [tex]HF[/tex]
- First calculate the number of moles of each compound using its molar mass and the mass that reacted completely:
[tex]13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{101.96gAl_{2}O_{3}}*\frac{1000g}{1kg}=131molesAl_{2}O_{3}[/tex]
[tex]55.4kgNaOH*\frac{1molNaOH}{40kgNaOH}*\frac{1000g}{1kg}=1385molesNaOH[/tex][tex]55.4kgHF*\frac{1molHF}{20kgHF}*\frac{1000g}{1kg}=2770molesHF[/tex]
- Divide the number of moles obtained between the stoichiometric coefficient of each compound in the chemical reaction:
[tex]Al_{2}O_{3}:\frac{131}{1}=131[/tex]
[tex]NaOH:\frac{1385}{6}=231[/tex]
[tex]HF:\frac{2770}{12}=231[/tex]
The [tex]Al_{2}O_{3}[/tex] is the limiting reagent because it has the smallest number.
3. Find the mass of cryolite produced:
[tex]13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{0.10196kgAl_{2}O_{3}}*\frac{2molesNa_{3}AlF_{6}}{1molAl_{2}O_{3}}*\frac{0.20994kgNa_{3}AlF_{6}}{1molNa_{3}AlF_{6}}=55.2kgNa_{3}AlF_{6}[/tex]
