[swati]

Thats exactly what I needed to see. Thanks.

Welcome .
But never expect spoon feeding . If someone gave you this answer in the starting then you won't use your brain . Now you understood the concept & you won't forget it easily . 

[Shea]

How many g of zinc are required to produced 50g of hydrogen when reacting zinc with sulphuric acid?

Did I do this right -->  Zn + H2SO4 -> 2H + ZnSO4

2 moles of H make 1 mole of Zn.

2g of H = 2 moles of H
50 g = 50 moles.

50 moles of H = 25 moles of Zn

25 moles of Zn = 1635 g.

1635 g of Zn * 25 = 40875 g

Did I mess up anywhere?

[Borek]

Did I mess up anywhere?

Yes, twice, but these are minor mistakes, you understand how to do the question.

Did I do this right -->  Zn + H2SO4 -> 2H + ZnSO4

H₂, not 2H. Luckily, mass is the same.

25 moles of Zn = 1635 g.

1635 g of Zn * 25 = 40875 g

You have already calculated mass of 25 moles of Zn above, second multiplication is a mistake.

[Shea]

So 1635 g of Zinc are required to produce 50g of hydrogen when reacting zinc with sulphuric acid?

[Borek]

Yes. 1622g if you want to be really precise - or, 1.6*10³g to be correct when it comes to significant digits.

[Shea]

Ok, I guess that sorta counts as me doing one for myself for once. 

Can you check my work on finding out how much sulphuric acid was needed for the previous question?


Zn + H2SO4 -> H2 + ZnSO4

2 moles of H = 1 mole of H2SO4

25 moles of H2SO4 = 2451.9625 g?

[swati]

Can you check my work on finding out how much sulphuric acid was needed for the previous question?

Zn + H2SO4 -> H2 + ZnSO4

2 moles of H = 1 mole of H2SO4

25 moles of H2SO4 = 2451.9625 g?

It is correct , but instead of 2 moles of H write 1 mole of H₂ .

Note : Mass of 1 mole of H₂ is also 2 g

[Borek]

You may check it by yourself - just download and install trial version of EBAS (see link in my signature), start the program, mark the reaction equation as written in your post and copy it to clipboard, press Ctrl-F to create reaction from clipboard content, and enter 50 as hydrogen mass.

Your result looks more or less correct, although it abuses significant digits. EBAS shows 2433 g (in output frame, below H₂SO₄ formula), but it uses much more accurate molar masses. To be really correct you should give answer as 2.4*10³ g.

[Shea]

I've been trying to use that, and, for the most part, I respect it a lot.  But what can it do for me if I can't even enter the proper equation?

Anyway, my teacher wants me to show work, so I have to do it myself.

I'm kinda getting the hang of it though.

[Borek]

what can it do for me if I can't even enter the proper equation?

Nothing.

Anyway, my teacher wants me to show work, so I have to do it myself.

And I am not proposing you to rely solely on the program, just to check the results. I won't help you for the next 8 hours, it is after midnight here and I am changing position to more horizontal

[jennielynn_1980]

There is a partial list of ions on this site:

http://chemtutor.com/compoun.htm

also some text books have lists of ions.  For example, Foundations of Chemistry 2nd editions by Toon and Ellis has a list of common ions.

Good luck

[Shea]

Thanks for the list.

Now, this question seems to be different from the ones yesterday.  How do I do it?

How many grams of fluorine are required to react with 100g of NaBr in the following reaction:

F2 + 2NaBr -> 2NaF + Br2

[Borek]

It is identical. Try to use ratios, as described on my page:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=ratio-proportions

or go for NaBr moles, F₂ moles - and convert them back to mass.

[Shea]

Do I have to do anything with the 2NaF?

[Borek]

Do I have to do anything with the 2NaF?

What for? You are asked only about reactants, amounts of products doesn't matter.