Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide.
Balance the equation:
Al2O3(s) + NaOH(l) + HF(g) -> Na3AlF6 + H2O(g)
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?
Which reactants will be in excess?
-Al2O3
-NaOH
-HF
What is the total mass of the excess reactants left over after the reaction is complete?
A long problem. What do you not understand about EACH part. Perhaps we can help you through it.
To balance the given chemical equation Al2O3(s) + NaOH(l) + HF(g) -> Na3AlF6 + H2O(g), we need to ensure that the number of atoms of each element is the same on both sides of the equation.
1. Start by balancing the aluminum (Al) atoms: There are 2 Al atoms on the left-hand side and 1 Al atom on the right-hand side. We can balance this by putting a coefficient of 2 in front of Na3AlF6:
Al2O3 + NaOH + HF -> 2Na3AlF6 + H2O
2. Next, balance the oxygen (O) atoms: There are 3 O atoms in Al2O3 and 1 O atom in H2O. To balance this, we need to multiply H2O by 3:
Al2O3 + NaOH + HF -> 2Na3AlF6 + 3H2O
3. Balance the sodium (Na) atoms: There are 3 Na atoms in Na3AlF6. We can balance this by putting a coefficient of 6 in front of NaOH:
Al2O3 + 6NaOH + HF -> 2Na3AlF6 + 3H2O
4. Finally, balance the hydrogen (H) atoms: There are 6 H atoms in 6NaOH and 6 H atoms in 3H2O. The equation is now balanced:
Al2O3 + 6NaOH + HF -> 2Na3AlF6 + 3H2O
Now that the equation is balanced, we can proceed to calculate the mass of cryolite produced.
To find the limiting reactant, we need to compare the number of moles of each reactant with their respective stoichiometric coefficients. Let's calculate the number of moles for each reactant:
1. Calculate moles of Al2O3:
Molar mass of Al2O3 = 102 g/mol
Moles of Al2O3 = mass of Al2O3 / molar mass of Al2O3
= 14.0 kg / 102 g/mol
= 137.25 mol (approximately)
2. Calculate moles of NaOH:
Molar mass of NaOH = 40 g/mol + 16 g/mol + 1 g/mol = 58 g/mol
Moles of NaOH = mass of NaOH / molar mass of NaOH
= 57.4 kg / 58 g/mol
= 988.28 mol (approximately)
3. Calculate moles of HF (note: gas):
Molar mass of HF = 1 g/mol + 19 g/mol = 20 g/mol
Moles of HF = mass of HF / molar mass of HF
= 57.4 kg / 20 g/mol
= 2870 mol (approximately)
Now, let's determine the limiting reactant. The stoichiometric coefficients tell us that the ratio of Al2O3 to NaOH to HF is 1:6:1. Therefore, to find the limiting reactant, we compare the mole ratio of Al2O3 to the other reactants:
Al2O3:NaOH = 1:6
Al2O3:HF = 1:1
Since the mole ratio of Al2O3 to NaOH is smaller than the mole ratio of Al2O3 to HF, NaOH is the limiting reactant.
To calculate the mass of cryolite produced, we need to use the stoichiometry from the balanced equation. The stoichiometric coefficient of Na3AlF6 is 2, meaning that for every 2 moles of Na3AlF6, we need 1 mole of Al2O3:
Moles of Na3AlF6 = Moles of Al2O3 / 1
= 137.25 mol
Now, calculate the mass of cryolite produced:
Molar mass of Na3AlF6 = 23 g/mol + 27 g/mol + (6 * 19 g/mol) = 210 g/mol
Mass of cryolite produced = Moles of Na3AlF6 * molar mass of Na3AlF6
= 137.25 mol * 210 g/mol
= 28,762.5 g
= 28.7625 kg
Therefore, 28.7625 kilograms of cryolite will be produced.
Now let's determine the mass of the excess reactants left over.
1. Excess NaOH:
To find the moles of excess NaOH, we subtract the moles of NaOH used (6 times the moles of Na3AlF6) from the initial moles of NaOH:
Moles of excess NaOH = Initial moles of NaOH - (6 * Moles of Na3AlF6)
= 988.28 mol - (6 * 137.25 mol)
= 158.53 mol
Now, calculate the mass of excess NaOH:
Mass of excess NaOH = Moles of excess NaOH * Molar mass of NaOH
= 158.53 mol * 58 g/mol
= 9192.74 g
= 9.19274 kg
2. Excess HF (note: gas):
To determine the mass of excess HF, we subtract the moles of HF used (the same as the moles of Na3AlF6) from the initial moles of HF:
Moles of excess HF = Initial moles of HF - Moles of Na3AlF6
= 2870 mol - 137.25 mol
= 2732.75 mol
Now, calculate the mass of excess HF:
Mass of excess HF = Moles of excess HF * Molar mass of HF
= 2732.75 mol * 20 g/mol
= 54,655 g
= 54.655 kg
Therefore, the mass of the excess reactants left over after the reaction is complete is approximately 9.19274 kilograms of NaOH and 54.655 kilograms of HF.