II-53. Cryolite, Na3 AlF6(s), an ore used in the production...

II-53. Cryolite, $\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)$, an ore used in the production of aluminum, can be synthesized by the reaction described by the equation $\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{NaOH}(l)+\mathrm{HF}(g) \stackrel{\text { High } T}{\longrightarrow}$ $\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad($ unbalanced $)$ (a) Balance this equation. (b) If $10.0$ kilograms of $\mathrm{Al}_{2} \mathrm{O}_{3}(s), 50.00$ kilograms of $\mathrm{NaOH}(l)$, and $50.0$ kilograms of $\mathrm{HF}(g)$ react completely, how many kilograms of cryolite will be produced? (c) Which reactants will be in excess and how many kilograms of each of these reactants will remain?

II-53. Cryolite, $\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)$, an ore used in the production of aluminum, can be synthesized by the reaction described by the equation
$\mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{NaOH}(l)+\mathrm{HF}(g) \stackrel{\text { High } T}{\longrightarrow}$
$\mathrm{Na}_{3} \mathrm{AlF}_{6}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad($ unbalanced $)$
(a) Balance this equation.
(b) If $10.0$ kilograms of $\mathrm{Al}_{2} \mathrm{O}_{3}(s), 50.00$ kilograms of $\mathrm{NaOH}(l)$, and $50.0$ kilograms of $\mathrm{HF}(g)$ react completely, how many kilograms of cryolite will be produced? (c) Which reactants will be in excess and how many kilograms of each of these reactants will remain?

Key Concepts

Yield Calculation

Yield calculation involves determining the amount of product that can be formed based on the limiting reactant. This process includes establishing the theoretical yield from stoichiometric ratios and adjusting for any excess reactants, which is essential for predicting actual production quantities.

Limiting Reactant

The concept of the limiting reactant is used to identify which reactant will be completely consumed first in a chemical reaction. This reactant dictates the maximum amount of product that can be formed, making it a central concept in practical yield calculations.

Mass-to-Mole Conversions

Mass-to-mole conversions are a key part of stoichiometric calculations. By using the molar mass of a substance, one can convert a given mass into the number of moles, which is necessary for applying the mole ratios of a balanced chemical equation.

Stoichiometry

Stoichiometry involves using a balanced chemical equation to calculate the quantitative relationships between reactants and products. It encompasses converting masses to moles, using mole ratios from the balanced equation, and converting back to mass to predict how much product will form from given amounts of reactants.

Balancing Chemical Equations

Balancing chemical equations is the process of ensuring that the number of atoms for every element is the same on both sides of the reaction. This reflects the law of conservation of mass and is essential for accurately determining the stoichiometric relationships between reactants and products.

Transcript

00:01 This question is fairly lengthy with multiple parts.

00:05 We have a reaction we have to first balance aluminum oxide reacting with sodium hydroxide and hf to produce cryolite and water.

00:15 To balance it, we've got one aluminum and two aluminums.

00:22 So if we put a two there, we'll now have two aluminums.

00:25 However, when we put a two there, that gives us six sodium.

00:28 So we need to put a six there.

00:30 So now we have six sodiums.

00:32 When we do that, when we did put this two here, that gave us 12 fluorines, so we need to put a 12 there.

00:40 So we have 12 fluorines.

00:42 That then gives us total of 18s.

00:45 We've got 12 times 1 hydrogens, that's 12 hydrogens, plus 6 times the hydrogens here.

00:52 That gives us 18 hydrogens.

00:57 So we need to put a 9 right there, so we get 18 hydrogens.

01:00 That now gave us nine oxygens.

01:03 So let's make sure we have nine oxygens.

01:05 Six times one plus three more is nine.

01:08 So it's finally balanced.

01:13 We have 10 kilograms of aluminum oxide.

01:17 We have 50 kilograms of sodium hydroxide.

01:20 And we have 50 kilograms of hf.

01:22 So we need to take all three of these amounts of the three reactants and calculate how much product can be produced with each of them, the product being cryopause.

01:33 The one that produces the least amount will be the limiting reactant, and that will be the amount of cryolite that we can produce.

01:40 So we'll start first with the 10 kilograms of aluminum oxide.

01:46 We'll convert them to grams by dividing by multiplying by a thousand.

01:51 Then when we have grams, we'll divide by the molar mass of al203 to produce moles of al203.

02:00 Then we'll use the stoichiometry of the balanced chemical reaction 1 mole al203 produces 2 moles of cryolite then when we know moles cryolite we'll multiply by the molar mass of cryolite which will be the mass of 3 moles sodium plus 1 mole aluminum and 6 moles fluorine or 209 .94 grams then we have grams we'll divide by a thousand to get kilograms and the 10 kilograms of aluminum oxide will produce 41 .2 kilograms of cryolite...

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