Question

How much naalo2 is required to produce 1. 00 kg of na3alf6? 6hf(g) + 3 naalo2(s) --> na3alf6(s) + 3h2o(l) + al2o3(s).

Answer 1

Considering the reaction stoichiometry,1171.43 grams of NaAlO₂ is required to produce 1.00 kg of Na₃AlF₆.

The balanced reaction is:

6 HF + 3 NaAlO₂ → Na₃AlF₆ + 3 H₂O + Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• HF: 6 moles
• NaAlO₂: 3 moles
• Na₃AlF₆: 1 mole
• H₂O: 3 moles
• Al₂O₃: 1 mole

The molar mass of each compound is:

• HF: 20 g/mole
• NaAlO₂: 82 g/mole
• Na₃AlF₆: 210 g/mole
• H₂O: 18 g/mole
• Al₂O₃: 102 g/mole

So, by reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:

• HF: 6 moles× 20 g/mole= 120 grams
• NaAlO₂: 3 moles× 82 g/mole= 246 grams
• Na₃AlF₆: 1 mole× 210 g/mole= 210 grams
• H₂O: 3 moles× 18 g/mole= 54 grams
• Al₂O₃: 1 mole× 102 g/mole= 102 grams

Then you can apply the following rule of three: if by stoichiometry 210 grams of Na₃AlF₆ are produced by 246 grams of NaAlO₂, 1000 g (1kg) of Na₃AlF₆ are produced by how much mass of NaAlO₂?

$mass of NaAlO_{2} =\frac{1000 grams of Na_{3} AlF_{6} x246 grams of NaAlO_{2}}{210 grams of Na_{3} AlF_{6}}$

mass of NaAlO₂= 1171.43 grams

Finally, 1171.43 grams of NaAlO₂ is required to produce 1.00 kg of Na₃AlF₆.

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