Question

5. (20 points) Cryolite (Na3AlF6) is used in the commercial production of aluminum from ore. Cryolite itself is produced by the following reaction: 6 NaOH + Al2O3 + 12 HF -> 2 Na3AlF6 + 9 H2O If 200 g of NaOH are reacted with 200 g of Al2O3 and excess HF: Molar masses: NaOH = 40.0 g/mole Al2O3 = 102 g/mole Na3AlF6 = 210 g/mole H2O = 18.0 g/mole A. How many grams of H2O are produced? B. Which reactant is the limiting reactant and which is in excess? Circle your answer below. NaOH limiting reactant in excess Al2O3 limiting reactant in excess C. How much of the excess reactant (not HF) remains after the limiting reactant is completely consumed? D. If the actual yield of H2O is 75.3 g, what is the percent yield?

          5. (20 points) Cryolite (Na3AlF6) is used in the commercial production of aluminum from ore. Cryolite itself is produced by the following reaction:
6 NaOH + Al2O3 + 12 HF -> 2 Na3AlF6 + 9 H2O
If 200 g of NaOH are reacted with 200 g of Al2O3 and excess HF:
Molar masses: NaOH = 40.0 g/mole Al2O3 = 102 g/mole
Na3AlF6 = 210 g/mole H2O = 18.0 g/mole
A. How many grams of H2O are produced?
B. Which reactant is the limiting reactant and which is in excess? Circle your answer below.
NaOH limiting reactant in excess
Al2O3 limiting reactant in excess
C. How much of the excess reactant (not HF) remains after the limiting reactant is completely consumed?
D. If the actual yield of H2O is 75.3 g, what is the percent yield?

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5. (20 points) Cryolite (Na3AlF6) is used in the commercial production of aluminum from ore. Cryolite itself is produced by the following reaction:
6 NaOH + Al2O3 + 12 HF -> 2 Na3AlF6 + 9 H2O
If 200 g of NaOH are reacted with 200 g of Al2O3 and excess HF:
Molar masses: NaOH = 40.0 g/mole Al2O3 = 102 g/mole
Na3AlF6 = 210 g/mole H2O = 18.0 g/mole
A. How many grams of H2O are produced?
B. Which reactant is the limiting reactant and which is in excess? Circle your answer below.
NaOH limiting reactant in excess
Al2O3 limiting reactant in excess
C. How much of the excess reactant (not HF) remains after the limiting reactant is completely consumed?
D. If the actual yield of H2O is 75.3 g, what is the percent yield?

Added by Leika L.

Chemistry Principles and Reactions

William L Masterton; Cecile N Hurley;… 7th Edition

Chapter 3

Instant Answer

Solved by Expert David Collins

03/01/2024

Step 1

The reactants and products do not balance, and some of the substances mentioned (like NaOOI and NLOH) do not exist or are not commonly used in chemistry. However, assuming that the equation is meant to be: \[ 6 \mathrm{NaOH}+\mathrm{NH}_3+12 \mathrm{HF} Show more…

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Transcript

00:01 We have our balanced chemical reaction where 6 moles of naoh react with al2o3 and 12 moles of hf to produce 2 moles of cryolite na3alf6 and 9h2o.

00:36 So if we have 200 grams of sodium hydroxide, we can convert that into moles using its molar mass, or one mole, naoh, is 40 grams.

01:04 And then we go from moles sodium hydroxide 6 to moles of cryolite 2.

01:15 Once we have moles of cryolite we can convert that into mass using its molar mass 1 1 mole na3alf6 is 210 grams.

01:41 We then perform a similar calculation with our 200 grams of aluminum oxide.

01:51 We convert it into moles using its molar mass.

01:54 One mole aluminum oxide is 102 grams.

02:04 Then from the stoichiometry we see that one mole aluminum oxide yields two moles of cryolite.

02:18 And then last of all we multiply by the molar mass of cryolite to get grams.

02:28 So one of these is going to produce a smaller amount of cryolite.

02:32 That will be known as the limitine reactant.

02:35 And the amount of cryolite the limitine reactant produces is all we're going to get.

02:41 So if all of the sodium hydroxide react, we get 350 grams cryolite and if all of the aluminum oxide reacts we would get 824 grams cryolite.

03:02 Well we're not going to get the 824 because we're going to run out of our limiting reactant sodium hydroxide.

03:13 Whoops! well that gives you the mass of cryolite but i guess they just want the mass of water...

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