Posted by Rachel on Saturday, October 6, 2012 at 1:36pm.
This is a limiting reagent problem. I know that because amounts are given for more than one reactant.
1. You need a balanced equation.
2. Convert 13.3 kg Al2O3 to mls.mols = grams/molar mass.
Do the same for NaOH.
Do the same for HF.
3. Using the coefficients in the balanced equation, convert mols Al2O3 to mols of the product.
Do the same for NaOH.
Do the same for HF.
4. It is more than likely that the three numbers will not be the same which means two of them are wrong. The correct value in limiting reagent problems is ALWAYS the smallest value and the reagent providing that number is the limiting reagent.
5. Convert the smallest number to grams of the product. g = mols x molar mass.
6. The other two reactants will be in excess. Using the coefficients again, and in the same manner as in step 3, convert mols of the limiting reagent to mols of one of the non-limiting reagents. Convert to grams and subtract from the initial grams. That will give you the amount of unreacted material at the end of the reaction.
7. Do the same for the other non-limiting reagent.
8. The problem asks for the total mass; therefore, add the amounts of th excess reagents to find the total.
This sounds long and involved but if you follow each step it will work. Post your work if you get stuck.
I forgot to input the equation. I balanced it and came up with: Al2O3(s)+6NaOH(l)+12HF(g) YIELDS 2Na3AlF6+9H20(g)
Then when I converted 13.3 kg Al2O3 to moles and got an answer of 7.6662 moles
After that I did the same for NaOH & HF and got: NaOH = 0.6848 moles, HF= 0.3426 moles.
* That is as far as I've gone, now I am confused. How do you convert mols of Al2O3 to mols of the product?
I have different numbers for mols. First let's do that. mol = g/molar mass.
mol Al2O3 = 13,300/101.96 = 130.44 mols.
mol NaOH = 58,400/40 = 1,460 mols.
mol HF = 58,400/20 = 2920 mols.
You convert to mols of Na3AlF6 this way using the coefficients in the balanced equation.
130.44 mol Al2O3 x (2 mol Na3AlF6/1 mol Al2O3) = 130.44 x (2/1) = 260.88 mols Na3AlF6.
1460 mol NaOH x (2 mol Na3AlF6/6 mol NaOH) = 1460 x (2/6) = 1460 x 1/3 = 486.7
2929 mol HF x (2 mol Na3AlF6/12 mols HF) = 2920 x (2/12) = 2920 x 1/6 = 486.7
Check these numbers carefully. Sometimes I punch the wrong keys.
Balanced equation of Al2O3 + HF