High school chem

[RichardJR]

Hi guys. I need some help with my homework. Here is the question and then I will tell you where I got stuck

 

What volume of a 3.75 M MgSO4 solution can be prepared with 271g of MgSO4?

 

Ok so I first calculated the molar mass of MgSO4 with the periodic table. Mg= 24.3. S= 32 and O4 = 16x4=65

 

TOTAL mass of MgSO4 is 120.3g and 271g of it is being used (so about 2.252x more than the initial mass. 2.252 x 120.3 = 271)

 

 

Sadly I am now stuck. Can someone give me a hint? I will then try to finish figuring it out and post my final answer. Thank you

 

Ok so I just looked at it and did 3.75 x 120.3 = 451.125. I then divided that by 271g to get 1.16646 etc which I bring down to 1.7L.

 

Or do I do 3.75M x 120.3 g = 451.125 / 2.252 (because 271g is 2.252 M of MgSO4) = 200.32 ML?

Edited July 17, 2010 by RichardJR

[Cap'n Refsmmat]

Here's what you need to do. If MgSO₄ has a molar mass of 120.4g, and 271g are being used, how many moles do you have? Calculate that first.

 

Next, consider the definition of molarity:

 

[math]\mbox{molarity} = \frac{\mbox{moles}}{\mbox{liters}}[/math]

 

You know the molarity you want - 3.75. You can calculate (like I just asked you to) the number of moles of solute you have. Solve the equation for liters and you're done.

[RichardJR]

Thank you for the quick reply. You seemed to have explained it pretty straight forward and it seems easy to understand. Hopefully that is enough for me =/

 

Molarity = Moles/Liters but since I don't have liters I need to do L=Moles/molarity I assume.

 

Therefore 271g of MgSO4 is 2.252 moles and so 2.252/3.75 = 0.6 L

[Cap'n Refsmmat]

Looks right to me.

[RichardJR]

Thank you for the help. I require a bit more help if anyone is still willing. I have these homework books and out of about 40 + questions I have been stuck on 6 (now 5)

 

I can't find any examples in my booklet to help me with this one. None relate, at least not from what I see... which is usually the problem.

 

You have to use an insecticide vaporizer with a 7.0 L capacity to treat your vine for an infestation of aphids. The manufacturer supplies a 3.5 M solution of S.O.S Nature, a natural insecticide, in an 850 ml format. You must spray your vine twice a week for two weeks with 1.25 L of solution diluted to 0.375 M for each application. How much of the concentrated solution do you need to prepare the complete treatment?

 

So what I've got so far.

 

Concentrated solution = Insecticide 3.5 M. I need to know how much of this I need to prepare.

 

1.25 L of solution must be diluted to 0.375 M EACH application (4 applications total). So 0.375 x4 = 1.5 M

 

Formula for dilution = c1v1=c2v2

 

Even knowing the formula, I have no idea what to do now..

Edited July 17, 2010 by RichardJR

[Cap'n Refsmmat]

Hmm. Here's what I'd do: Figure out how many moles you'd have to do to apply the insecticide. 1.25L at 0.375M twice a week for two weeks. You can calculate (using the formula I gave above) how many moles of insecticide that is.

 

Then, you can figure out how much concentrated solution you'd need the same way. You know the number of moles you need, and you know the molarity, so you can find the number of liters you need.

[RichardJR]

1.25 / 0.375 = .3

 

0.3 x 4 (4 applications) = 1.2

3.5 / 1.2 = 2.92L needed of the concentrated solution. Still not sure about this answer though.

Edited July 17, 2010 by RichardJR

[Cap'n Refsmmat]

Check your first step there. 1.25 / 0.375 = 0.3? And I don't think you want to be doing division...

 

Make sure you plug things into the correct spots in the equation I gave in post #2.

[RichardJR]

The manufacturer supplies a 3.5 M solution of S.O.S Nature, a natural insecticide, in an 850 ml format. You must spray your vine twice a week for two weeks with 1.25 L of solution diluted to 0.375 M for each application. How much of the concentrated solution do you need to prepare the complete treatment?

 

1.25L x 0.375 = 0.47 M

 

0.47 x 4 applications= 1.88 M

 

1.88 M / 3.5 = 0.54 L

 

Then I calculated it another way. 0.47 / 3.5 = 0.134

 

4 applications = 0.134 x 4 = 0.536 = 0.54 L

 

So you need 0.54L of the solution. Is that better or still completely off? Seems right to me this time

[Cap'n Refsmmat]

Sounds right to me.

[RichardJR]

Thank you. I only have 4 more questions unanswered.

 

When sodium (Na) reacts with water, it produces sodium hydroxide (NaOH) according to the following equation.

 

H2O + Na ---> NaOH + H2

Determine the number of moles of sodium hydroxide produced when 9 ml of water react according to the equation above. Note: The mass of 1 millilitre of water is 1 gram.

 

With the periodic table I know H2O molar mass is 1g (H) x 2 = 2g + O = 16 G = 18g total

 

So 1 mole of H2O is 18g. Now Sodium hydroxide. Na= 23G, O = 16 and H = 1 which = 40g/mole of NaOH

 

So 1 mole of H2O is 18g and it wants to know the number of moles of sodium hydroxide produced when 9g (half a mole of h2o) of h2o reacts with Na if I understand correctly. Can I get a bit of help to get me further along the way? Thanks

 

edit: may as well take a shot no matter how bad it is. So basically half a mole of H2O is 9g. Na = 23, so 23 + 9g = 32g of NaOH. 1 mole of NaOH is 40g. So 32/40 = 0.8 M of NaOH is produced.

Edited July 18, 2010 by RichardJR

[Cap'n Refsmmat]

You have this chemical equation:

 

[ce]H2O + Na -> NaOH + H2[/ce]

 

The first thing you need to do is balance that. You won't get the right answer until it's a balanced equation.

[RichardJR]

H2O + Na ---> NaOH + H2

 

Left side 2 atoms of H, 1 atom of O, 1 atom of Na

 

Right side 3 atoms of H, 1 atom of O, 1 atom of Na

 

Balanced equation should then be

 

2 H2O + 2 Na ---> 2 NaOH +H2

 

More clueless than before but I'll try like my first time. 9g of H2O + 2 NA = 23g x2 = 46 + 9 = 54g

 

One mole of NaOH = 40g. So 54/40 = 1.35 M

Edited July 18, 2010 by RichardJR

[Cap'n Refsmmat]

Okay:

 

[ce]2H2O + 2Na -> 2NaOH + H2[/ce]

 

Easy way to do this: 9g of [ce]H2O[/ce] is half a mole, because you discovered it's 18g/mol, yes?

 

So we have 0.5 moles. According to our chemical equation above, 2 moles of NaOH are produced for every 2 moles of [ce]H2O[/ce] -- that's a 1 to 1 ratio. So, put in 0.5mol water and you get 0.5mol NaOH.

 

How to do this in harder problems:

 

There's a method called dimensional analysis that works very well.

 

[math]\frac{0.5 \mbox{mol}}{ } \times \frac{18 \mbox{g} }{1 \mbox{mol} } \times \frac{2\mbox{mol water}}{2 \mbox{mol NaOH}} = 9 \mbox{g NaOH}[/math]

 

Here's a tutorial:

 

http://www.chembuddy...sional-analysis

[RichardJR]

Hey thanks a lot. I have 3 more questions but they're all basically the same so I'll ask one and figure out the rest myself.

 

How many moles of ammonium chloride (NH4Cl) will be produced from 84.5g of iron chloride (FeCl2) according to the reaction below?

 

FeCl2 + NH4OH ---> Fe(OH)2 + NH4Cl

BALANCED

 

FeCl2 + 2 NH4OH ---> Fe(OH)2 + 2 NH4Cl

 

molar mass of NH4Cl = 14g + 4g + 35.5g= 1 mole of NH4Cl is 53.5G

molar mass of FeCl2 = 55.9g+70g = 1 mole of FeCl2 is 125.9g

 

Now from the previous example we can see the ratio is 1:2. 1 mole of FeCL2 makes 2 moles of NH4Cl.

 

So 84.5g of FeCl2 = 0.67g M of FeCl2. Now since 1 M of FeCl2 makes 2 moles of NH4Cl, do I not just times 0.67g x 2? So 84.5g of iron chloride makes 1.34 M of ammonium chloride.

[Cap'n Refsmmat]

Sounds good.

 

One note: capital M usually denotes molarity. To prevent confusion, I'd use "mol" for "8.2 mol" or whatever.

[RichardJR]

One last question (actually has followups in a/b/c fashion) and then I can send these homework books (3 total) to be corrected. I'll post the results when I get them back!

 

A student has forgotten to identify the four flasks containing liquids that he needs for a project. The labels are on the table and they are marked as follows:

 

0.05 M Pb(OH)2

0.01 M H2CO3

0.01 M PbSO4

C5H10

 

Match each label with one of the categories. Acid/base/salt/non-electrolyte

 

I classified Pb(OH)2 as a base because it has OH in it.

I classified PbSO4 as a salt because it contains anions/cations other than OH or H.

I classified 0.01 M H2CO3 as a non electrolyte because it's a weak acid.

I classified C5H10 as an acid because it was the only option left. I have no idea what C5H10 is even supposed to represent.

I think I have this correct but it's the followup question that I can't answer.

 

Suggest a test that would help you identify the non-electrolyte liquid conclusively. Explain. What label corresponds to this solution?

 

 

Then

 

Suggest a method that would enable you to distinguish the three other substances. Indicate the results obtained for each case and the label which corresponds to each solution.

 

So for this one I wrote. You can distinguish them by using litmus paper. A strong acid turns BLUE litmus paper RED. A strong base turns RED litmus paper BLUE.

 

Salt - No idea. I need to know a method that can distinguish it from acids and bases. If you put a salt in water, it dissolves. Could this be an accurate method?

 

As for the first question, I have absolutely no idea.

[Cap'n Refsmmat]

I classified 0.01 M H2CO3 as a non electrolyte because it's a weak acid.

I classified C5H10 as an acid because it was the only option left. I have no idea what C5H10 is even supposed to represent.

Perhaps you could label the weak acid as an acid, rather than as a non-electrolyte? That might make more sense.

 

Suggest a test that would help you identify the non-electrolyte liquid conclusively. Explain. What label corresponds to this solution?

[ce]C5H10[/ce] is a hydrocarbon - pentene. I'm not sure what they're looking for here, but hydrocarbons are combustible, so that's one idea.

 

Suggest a method that would enable you to distinguish the three other substances. Indicate the results obtained for each case and the label which corresponds to each solution.

 

So for this one I wrote. You can distinguish them by using litmus paper. A strong acid turns BLUE litmus paper RED. A strong base turns RED litmus paper BLUE.

 

Salt - No idea. I need to know a method that can distinguish it from acids and bases. If you put a salt in water, it dissolves. Could this be an accurate method?

 

As for the first question, I have absolutely no idea.

 

Acids and bases change the litmus paper, yes. Other things, like salts, don't. That should be all you need.