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KatRina [158]
3 years ago
7

Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f

or 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?
Chemistry
1 answer:
Korvikt [17]3 years ago
3 0

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}=1385molesNaOH55.4kgHF*\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}

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