If 16.3 kilograms of Al2O3(s), 60.4 kilograms of NaOH(l), and 60.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

Do the math

To determine the amount of cryolite produced in this reaction, we need to first identify the balanced chemical equation for the reaction between aluminum oxide (Al2O3), sodium hydroxide (NaOH), and hydrogen fluoride (HF).

The balanced chemical equation for this reaction is:

2 Al2O3(s) + 6 NaOH(l) + 12 HF(g) → 2 Na3AlF6(s) + 6 H2O(l)

According to the balanced equation, 2 moles of Al2O3 reacts with 6 moles of NaOH and 12 moles of HF to produce 2 moles of Na3AlF6.

To find the number of moles of Al2O3, NaOH, and HF, we can use their respective molar masses. The molar mass of Al2O3 is 101.96 g/mol, the molar mass of NaOH is 39.997 g/mol, and the molar mass of HF is 20.006 g/mol.

Now, let's calculate the number of moles of each substance:

Number of moles of Al2O3 = (mass of Al2O3) / (molar mass of Al2O3)
= 16.3 kg / (101.96 g/mol)
= 0.1599 mol

Number of moles of NaOH = (mass of NaOH) / (molar mass of NaOH)
= 60.4 kg / (39.997 g/mol)
= 1.5104 mol

Number of moles of HF = (mass of HF) / (molar mass of HF)
= 60.4 kg / (20.006 g/mol)
= 3.0194 mol

Since the reaction stoichiometry tells us that 2 moles of Al2O3 produce 2 moles of Na3AlF6, we can use the mole ratio to find the number of moles of Na3AlF6 produced.

Number of moles of Na3AlF6 = (Number of moles of Al2O3) x (2 moles of Na3AlF6 / 2 moles of Al2O3)
= 0.1599 mol x 1
= 0.1599 mol

To convert the number of moles of Na3AlF6 to kilograms, we can use its molar mass, which is 209.942 g/mol.

Mass of Na3AlF6 = (Number of moles of Na3AlF6) x (molar mass of Na3AlF6)
= 0.1599 mol x 209.942 g/mol
= 33.558 kg

Therefore, approximately 33.558 kilograms of cryolite will be produced when 16.3 kilograms of Al2O3, 60.4 kilograms of NaOH, and 60.4 kilograms of HF react completely.