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

Answer:

$55.2kgNa_{3}AlF_{6}$

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:

$13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{101.96gAl_{2}O_{3}}*\frac{1000g}{1kg}=131molesAl_{2}O_{3}$

$55.4kgNaOH*\frac{1molNaOH}{40kgNaOH}*\frac{1000g}{1kg}=1385molesNaOH$$55.4kgHF*\frac{1molHF}{20kgHF}*\frac{1000g}{1kg}=2770molesHF$

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

$Al_{2}O_{3}:\frac{131}{1}=131$

$NaOH:\frac{1385}{6}=231$

$HF:\frac{2770}{12}=231$

The $Al_{2}O_{3}$ is the limiting reagent because it has the smallest number.

3. Find the mass of cryolite produced:

$13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{0.10196kgAl_{2}O_{3}}*\frac{2molesNa_{3}AlF_{6}}{1molAl_{2}O_{3}}*\frac{0.20994kgNa_{3}AlF_{6}}{1molNa_{3}AlF_{6}}=55.2kgNa_{3}AlF_{6}$