* If 13.3 kilograms of Al2O3(s), 58.4 kilograms of NaOH(l), and 58.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?
* Which reactants will be in excess?
*What is the total mass of the excess reactants left over after the reaction is complete?
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
If 13.4 kilograms of Al2O3(s), 54.4 kilograms of NaOH(l), and 54.4 kilograms of HF(g) react completely, how many kilograms of cryolite will be produced?
To solve this problem, we need to analyze the balanced chemical equation and use stoichiometry to determine the amount of cryolite produced, as well as the excess reactants.
The balanced chemical equation for the reaction is:
2 Al2O3(s) + 6 NaOH(l) + 12 HF(g) → 2 Na3AlF6(s) + 6 H2O(l)
1. Determine the limiting reactant:
To find the limiting reactant, we need to calculate the number of moles of each reactant and compare them based on the stoichiometry of the balanced equation.
First, let's calculate the number of moles of each reactant:
- Al2O3(s): 13.3 kilograms * (1 mole / 101.96 grams) = 0.130 moles
- NaOH(l): 58.4 kilograms * (1 mole / 39.997 grams) = 1460 moles
- HF(g): 58.4 kilograms * (1 mole / 20.01 grams) = 2918 moles
Next, let's compare the moles of each reactant to the stoichiometry of the balanced equation:
- Al2O3(s): 0.130 moles * (2 moles Na3AlF6 / 2 moles Al2O3) = 0.130 moles Na3AlF6
- NaOH(l): 1460 moles * (2 moles Na3AlF6 / 6 moles NaOH) = 486.7 moles Na3AlF6
- HF(g): 2918 moles * (2 moles Na3AlF6 / 12 moles HF) = 486.3 moles Na3AlF6
From the calculations, we can see that NaOH(l) and HF(g) will produce slightly more moles of Na3AlF6 than Al2O3(s). Therefore, Al2O3(s) is the limiting reactant.
2. Determine the amount of cryolite produced:
Since the balanced equation tells us that 2 moles of Na3AlF6 are obtained from 2 moles of Al2O3, we can use stoichiometry to calculate the moles of Na3AlF6 produced from 0.130 moles of Al2O3:
0.130 moles Al2O3 * (2 moles Na3AlF6 / 2 moles Al2O3) = 0.130 moles Na3AlF6
Now, we can convert the moles of Na3AlF6 into kilograms:
0.130 moles Na3AlF6 * (209.94 grams / 1 mole) * (1 kilogram / 1000 grams) = 0.0273 kilograms Na3AlF6
Therefore, 0.0273 kilograms of cryolite will be produced.
3. Determine the excess reactants and the total mass left over:
The excess reactants are the ones that are not fully consumed in the reaction.
- NaOH(l) excess:
To determine the mass of excess NaOH(l), we need to determine how much NaOH(l) is left after it reacts with the limiting reactant (Al2O3). We can use stoichiometry to calculate the moles of NaOH(l) reacted with Al2O3:
0.130 moles Al2O3 * (6 moles NaOH / 2 moles Al2O3) = 0.390 moles NaOH(l)
The initial moles of NaOH(l) were 1460. Subtracting the moles reacted with Al2O3, we get:
1460 moles NaOH(l) - 0.390 moles NaOH(l) = 1459.610 moles NaOH(l)
Now, we can convert the moles of NaOH(l) into kilograms:
1459.610 moles NaOH(l) * (39.997 grams / 1 mole) * (1 kilogram / 1000 grams) = 58.383 kilograms NaOH(l)
Therefore, the excess reactant is NaOH, with a total mass of 58.383 kilograms left over.
- HF(g) excess:
Similarly, to determine the mass of excess HF(g), we need to determine how much HF(g) is left after it reacts with the limiting reactant (Al2O3). Using stoichiometry, we can calculate the moles of HF(g) reacted with Al2O3:
0.130 moles Al2O3 * (12 moles HF / 2 moles Al2O3) = 0.780 moles HF(g)
The initial moles of HF(g) were 2918. Subtracting the moles reacted with Al2O3, we get:
2918 moles HF(g) - 0.780 moles HF(g) = 2917.220 moles HF(g)
Now, we can convert the moles of HF(g) into kilograms:
2917.220 moles HF(g) * (20.01 grams / 1 mole) * (1 kilogram / 1000 grams) = 58.360 kilograms HF(g)
Therefore, the excess reactant is HF, with a total mass of 58.360 kilograms left over.
To summarize:
- The amount of cryolite produced is 0.0273 kilograms.
- The excess reactants are NaOH(l) with a total mass of 58.383 kilograms and HF(g) with a total mass of 58.360 kilograms left over.
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.