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Chemistry 7 Online
OpenStudy (anonymous):

Can someone help me with Stoichiometry?

OpenStudy (anonymous):

How many moles of aluminum metal would be needed to produce 35.6 grams of aluminum oxide in the following synthesis reaction? Al + O2 → Al2O3 there is an example of a problem.

OpenStudy (joannablackwelder):

Sure. First we need to balance the reaction. What goes in must come out. So we need 4Al + 3O2 -> 2AlO3. Following me so far?

OpenStudy (joannablackwelder):

I mean 2Al2O3.

OpenStudy (anonymous):

Yes :)

OpenStudy (joannablackwelder):

Great! So now we need to take what we have been given and use ratios to get what we want. We have 35.6 g of Al2O3. We can find out how many moles that is by using the molar mass.

OpenStudy (anonymous):

okay. gotcha. So keep going and explain molar mass to me..

OpenStudy (joannablackwelder):

Al is 26.9815 g/mol and O is 15.9994 g/mol (from the periodic table). 2 Als and 3 Os add up to 101.9612 g/mol.

OpenStudy (joannablackwelder):

The molar mass of Al2O3.

OpenStudy (anonymous):

you lost me on the second part.

OpenStudy (anonymous):

the 2 Als and Os. how did you get that

OpenStudy (joannablackwelder):

Because there are 2 Aluminums and 3 Oxygens in one mol of Al2O3.

OpenStudy (anonymous):

OH OKAY. that makes sense. okay. So how am i going to write out my equation.

OpenStudy (joannablackwelder):

Ok. Start with 35.6 g of Al2O3 * (1 mol Al2O3/101.9612 g Al2O3) = 0.349 mol Al2O3.

OpenStudy (anonymous):

okay and you got the 101.9612 from adding together the 26.9815 and the 15.999. right?

OpenStudy (joannablackwelder):

Well, adding 2(26.9815) + 3(15.9994).

OpenStudy (anonymous):

oh okay. wow this is confusing.

OpenStudy (joannablackwelder):

It should get easier with practice. :)

OpenStudy (anonymous):

okay. so was that all to it?

OpenStudy (anonymous):

of*

OpenStudy (joannablackwelder):

No, not yet.

OpenStudy (joannablackwelder):

Did you see how I converted from grams of Al2O3 to moles of Al2O3?

OpenStudy (anonymous):

uhm.. Sortof.. not really..>__<

OpenStudy (joannablackwelder):

Ok. This would be much easier to explain on a whiteboard. I can give you a free lesson if you want.

OpenStudy (anonymous):

lol okay

OpenStudy (anonymous):

Just use the draw button or something or draw it out and attach it or something. D:

OpenStudy (anonymous):

Something.

OpenStudy (joannablackwelder):

I am on an iPad, so I don't have a draw button. :/

OpenStudy (anonymous):

darn. okay, uhm... lets do this. I have another practice problem. Ill post that. and you don't have to tell me the answers, just guide me through it. and if i'm wrong, explain why. I'm homeschooled so I don't really have a teacher. and I don't have a test for this or anything yet, but I still want to understand it.

OpenStudy (joannablackwelder):

That is cool. I'm glad you want to understand it. I will try to break it down.

OpenStudy (anonymous):

How many grams of sodium metal are needed to react completely with 45.6 grams of water? okay. there's the problem. I need to make an equation first. right?

OpenStudy (joannablackwelder):

Ok, this problem has a little easier calculations. So, basically, the idea is that we can't convert from grams of one component to grams of another component. But we can figure out their mole ratios.

OpenStudy (joannablackwelder):

So, we need to convert from grams of water to moles of water to moles of sodium to grams of sodium.

OpenStudy (joannablackwelder):

Do you see the overall idea?

OpenStudy (anonymous):

I think so. I think i already took notes on how to do that awhile back. one second.

OpenStudy (joannablackwelder):

To convert grams of water to moles of water we need the molar mass of water.

OpenStudy (anonymous):

okay. so.. this is just me guessing. We need to put the molar mass of Hydrogen and Oxygen together. Right?

OpenStudy (joannablackwelder):

Right, but there are 2 Hydrogens for every 1 Oxygen in water (H2O), so we need to do 2(molar mass of Hydrogen) + 1(molar mass of Oxygen).

OpenStudy (joannablackwelder):

Try to calculate it and tell me what you get.

OpenStudy (anonymous):

So 2 (1.0079) + 1 (15.999)

OpenStudy (joannablackwelder):

Yes, perfect.

OpenStudy (anonymous):

= 18.0148

OpenStudy (joannablackwelder):

Great. So now we can calculate the moles of water by multiplying 45.6 g water by (1 mol water/18.0148 grams water) to cancel out grams of water and leave us with moles of water.

OpenStudy (joannablackwelder):

So, we have 2.53 moles of water.

OpenStudy (anonymous):

Right. thats what I got.

OpenStudy (anonymous):

And thats all...?

OpenStudy (joannablackwelder):

Now we need to look at the reaction of sodium and water to find their mole ratios.

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

so sodium is 22.990

OpenStudy (joannablackwelder):

True but first we need to look at the reaction. Na + H2O -> H2 + Na2O

OpenStudy (anonymous):

okay. so that is balanced.

OpenStudy (joannablackwelder):

No, we need 2 Sodiums in the reactants to balance it.

OpenStudy (joannablackwelder):

2Na + H2O -> H2 + Na2O.

OpenStudy (anonymous):

but doesnt the 2 on the other side just go with the O?

OpenStudy (anonymous):

there isnt 2 Na..

OpenStudy (joannablackwelder):

This means for every 2 moles of Na and 1 mole of H2O, we will make 1 mole of H2 and 1 mole of Na2O.

OpenStudy (anonymous):

you lost me. I'm so sorry. D:

OpenStudy (joannablackwelder):

No, the 2 behind the Na means 2 Nas bonded to 1 O.

OpenStudy (anonymous):

okay

OpenStudy (joannablackwelder):

This is very difficult to explain since I don't know any of your background and I can only type. :(

OpenStudy (anonymous):

well. pretty much. i'm a junior in high school and I'm horrible at math :D really anything helps. Explain it for the stupid. It helps.

OpenStudy (joannablackwelder):

I was just assuming that you understand the notation of chemical formulas.

OpenStudy (anonymous):

Uh well. okay. lets go with i understand the basics. I had some problems balancing equations.

OpenStudy (joannablackwelder):

Ok, so do you understand why I need to balance the formula as 2Na + H20 -> H2 + Na2O?

OpenStudy (anonymous):

No. Because i see it as. you have 1 Na on one side and now you have 2 Na on the other side.

OpenStudy (anonymous):

Thats how i see it. Feel free to explain why i'm wrong. haha

OpenStudy (joannablackwelder):

Na2O means 2 Nas and 1 O.

OpenStudy (anonymous):

Okay so the 2 goes with the Na. not the O.

OpenStudy (joannablackwelder):

This is an ionic compound. Na has a charge of +1 and O has a charge of -2. You need 2 Nas to balance out the O.

OpenStudy (anonymous):

okay. I sortof get it.

OpenStudy (anonymous):

So pretty much all you did was you balanced the equation

OpenStudy (joannablackwelder):

Right. So now, using the coefficients of the reaction, I can say that for every 2 moles of Na and 1 mole of H2O I can make 1 mole of H2 and 1 mole of Na2O.

OpenStudy (anonymous):

okay. i get that.

OpenStudy (joannablackwelder):

So we made it to 2.53 moles of water last time. We can convert to moles of Na by multiplying by (2 moles of Na/1 mole of H2O).

OpenStudy (joannablackwelder):

So we have 5.06 moles of Na.

OpenStudy (anonymous):

wait so my equation is 2.53 --> 2 mol Na / 1 h2O

OpenStudy (anonymous):

right?

OpenStudy (joannablackwelder):

No, 2.53 moles of H2O * (2 mol Na/1 mol H2O) = 5.06 moles Na

OpenStudy (anonymous):

okay thats what i got,

OpenStudy (joannablackwelder):

Great. Now we can use the molar mass of Na you mentioned earlier of 22.9898 g / mol. 5.06 moles * (22.9898 grams of Na/1 mol Na) = 116 g Na.

OpenStudy (anonymous):

okay. uhm. well. The steps it shows. Obviously do not make sense to me. That's why i'm asking. lol but. the lesson says 45.6g H2O x 1 mol H2O / 18.01g H2O x 2 mol Na / 2 mol H2O x 22.989 g Na / 1 mol Na = 58.2g Na

OpenStudy (joannablackwelder):

Sorry, I had to shut down. That is rright. I am sorry. I accidentally looked at the wrong equation. The actual reaction is 2Na + 2H2O -> H2 + 2NaOH.

OpenStudy (joannablackwelder):

So, if you change the molar ratio, you get what your book got. Sorry again...

OpenStudy (anonymous):

no big deal :)

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