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?

Answer 1

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.

Answer 2

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