If 14.0 kilograms of Al2O3(s), 57.4 kilograms of NaOh(l), and 57.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?

Al2O3 + 6NaOH + 12HF ==>2Na3AlF6 + 9H2O

Check that to make sure it is balanced.

This is a three-some limiting reagent problem. I work these by converting mols of the starting reagent to mols of the product. The smaller one wins. And we can save time by not changing the kg to grams.
Change 14.0 kg/kgmols for Al2O3. That's 14.0/101.96 = 0.137

Do the same for NaOH and HF.

Using the coefficients in the balanced equation, convert these kg-moles into kg moles of the product.
You will get different values for the product and all can't be right. The smaller number is the one you choose. Then change that to kg by kg-moles x molar mass of the product.
You should obtain in the neighborhood of 60 g if I didn't make a math error.

I should have said "the smallest number is the one to choose."

To determine the number of kilograms of cryolite produced, we first need to identify the balanced chemical equation for the reaction between Al2O3(s), NaOH(l), and HF(g).

The balanced chemical equation for the reaction is as follows:
Al2O3(s) + 6NaOH(l) + 12HF(g) → 2Na3AlF6(aq) + 3H2O(l)

From the balanced equation, we can see that every 1 mole of Al2O3 reacts with 6 moles of NaOH and 12 moles of HF to produce 2 moles of Na3AlF6.

To determine the molar masses of the reactants and products:
- The molar mass of Al2O3 is 101.96 g/mol.
- The molar mass of NaOH is 39.997 g/mol.
- The molar mass of HF is 20.01 g/mol.
- The molar mass of Na3AlF6 is 208.98 g/mol.

Next, we need to calculate the number of moles of Al2O3, NaOH, and HF in the given masses:
- Moles of Al2O3 = mass of Al2O3 / molar mass of Al2O3
- Moles of NaOH = mass of NaOH / molar mass of NaOH
- Moles of HF = mass of HF / molar mass of HF

Finally, we calculate the number of moles of Na3AlF6 produced:
- Moles of Na3AlF6 = (moles of Al2O3) × (2 moles of Na3AlF6 / 1 mole of Al2O3)

To find the mass of Na3AlF6 produced, we multiply the number of moles of Na3AlF6 by the molar mass of Na3AlF6:
- Mass of Na3AlF6 = (moles of Na3AlF6) × (molar mass of Na3AlF6)

Let's calculate the results step-by-step:
1. Moles of Al2O3:
- Moles of Al2O3 = 14.0 kg / 101.96 g/mol

2. Moles of NaOH:
- Moles of NaOH = 57.4 kg / 39.997 g/mol

3. Moles of HF:
- Moles of HF = 57.4 kg / 20.01 g/mol

4. Moles of Na3AlF6:
- Moles of Na3AlF6 = (moles of Al2O3) × (2 moles of Na3AlF6 / 1 mole of Al2O3)

5. Mass of Na3AlF6:
- Mass of Na3AlF6 = (moles of Na3AlF6) × (molar mass of Na3AlF6)

By performing these calculations, you will find the number of kilograms of cryolite produced.

To determine the number of kilograms of cryolite (Na3AlF6) produced, we need to first determine the limiting reactant among Al2O3(s), NaOH(l), and HF(g).

To do this, we compare the moles of each reactant to the stoichiometric coefficients in the balanced chemical equation:

2 Al2O3 + 6 NaOH + 12 HF ⟶ 2 Na3AlF6 + 9 H2O.

Let's calculate the moles of each reactant:

Moles of Al2O3 = mass of Al2O3 / molar mass of Al2O3
= 14.0 kg / (2 * (26.98 g/mol + 3 * 16.00 g/mol))
= 0.1348 mol

Moles of NaOH = mass of NaOH / molar mass of NaOH
= 57.4 kg / (22.99 g/mol + 16.00 g/mol + 1.01 g/mol)
= 2.0672 mol

Moles of HF = mass of HF / molar mass of HF
= 57.4 kg / (19.00 g/mol)
= 3.0211 mol

Using the balanced chemical equation, the stoichiometric ratio of Al2O3, NaOH, and HF to cryolite is 2:6:2 (in terms of moles of Al2O3:NaOH:HF).

To determine the limiting reactant, we compare the ratio of moles actually present to the stoichiometric ratio.

For Al2O3: 0.1348 mol / 2 = 0.0674 mol
For NaOH: 2.0672 mol / 6 = 0.3445 mol
For HF: 3.0211 mol / 2 = 1.5106 mol

From these calculations, we can see that Al2O3 is the limiting reactant because it has the smallest mole ratio compared to the stoichiometric ratio.

Now, let's calculate the moles of cryolite produced from the limiting reactant:

Moles of Na3AlF6 produced = moles of Al2O3 * (2/2) = 0.0674 mol

Finally, we can convert the moles of cryolite to kilograms:

Mass of Na3AlF6 = moles of Na3AlF6 * molar mass of Na3AlF6
= 0.0674 mol * (3 * (22.99 g/mol + 26.98 g/mol + 19.00 g/mol))
= 26.02 g

Therefore, the mass of cryolite produced is 26.02 kilograms.