## anonymous one year ago Gas calculation help?

1. anonymous

@iambatman you seem fun to work with, could you give me some assistance?

2. anonymous

hmm, @Abhisar you'll be my second choice, any time to help with simple a simple mole ratio? c:

3. anonymous

I just need help turning this Mg(s) + 2 HCl(aq) -> H2(g) + MgCl2(aq) into a mole ratio between H2/Mg 0.034 grams of Mg 0.0014 moles H2

4. anonymous

hmm, @Nnesha ?

5. Photon336

0.034 grams of Mg is what you have, and 0.0014 moles of H2 are produced? what are you being asked to do? and what information does the question give you

6. anonymous

Let's see, this is the question "If magnesium was the limiting reactant in this lab, calculate the theoretical yield of the gaseous product. Show all steps of your calculation."

7. anonymous

PV = nRT P = partial pressure H2 = 1.07 atm V = volume = 30 ml = 0.030 L n = moles = ? moles R = gas constant = 0.082057 L atm moles^-1 k^-1 T = temp in Kelvin = 24 deg C + 273.15 = 297.15 K n = PV / RT = 1.07 atm * 0.032 L / (0.082057 * 297.15 K) = 0.0014 moles H2 thiiiis might help

8. anonymous

(0.034g /24.03 Mg) * (2.02 g/moles H2) = 0.0028g I have this, but I just need a mole ratio with H2/Mg

9. Photon336

ok, so you know magnesium is the limiting reagent so you set it up like this: you need to figure out how much H2 was produced based off the stoichiometry this is theoretical yield. First find how many moles of magnesium you have $Grams magnesium* \frac{ 1 mol }{ 24.0g}$ then multiply this by the molar ratio and that's how much hydrogen gas you get. $Moles of magnesium * (\frac{ hydrogen gas }{ magnesium } = moles h2$ what you have is correct; the molar ratio of magnesium to hydrogen gas in that equation is 1:1

10. Photon336

(0.034g /24.03 Mg) * (2.02 g/moles H2) = 0.0028g I have this, but I just need a mole ratio with H2/Mg so the number of moles of magnesium = #moles of hydrogen gas again what you did is correct the molar ratio is 1:1

11. anonymous

thank you very much Photon c: I may need your help some other time, I'll fan you c:

12. Photon336

Yeah sure no problem

13. Photon336

FYI when they say theoretical they mean what you get from the stoichiometry

14. Photon336

Sometimes theoretical may not equal to what you get when you do the experiment; actually... if it is that's a pretty damned good experiment.. side reactions, and other factors can reduce your experimental value.

15. anonymous

Well, I'll pretend that this is a good experiment xD it's all virtual now, would be nice to see what this was actually like in person xD

16. anonymous

actually, I got this "Using grams of Mg used in the lab, calculate theoretical yield of the moles of Hydrogen gas. Use the mole ratio between H2/Mg to finish your work. Molar mass is not used here. " from my teacher. doing what I did, it basically came out to If magnesium was the limiting reactant in this lab, calculate the theoretical yield of the gaseous product. Show all steps of your calculation. (0.034g /24.03 Mg) * (2.02 g/moles H2) = 0.0028g = Actual yield Ratio of H2/Mg = 1:1 (0.034g /24.03 Mg) * (1 mole H2/1 mole Mg) = 0.0014 moles = Theoretical Yield Determine the percent yield of this reaction, showing all steps of your calculation. % yield = actual yield / theoretical yield x 100 / 1 0.0028 /0.0014 = 2 * 100/1 = 200% which seems pretty wrong, it was 0.0014/0.0028 at first but I found out that (0.034g /24.03 Mg) * (2.02 g/moles H2) = 0.0028g was the actual yield and (0.034g /24.03 Mg) * (1 mole H2/1 mole Mg) = 0.0014 moles would be the theoretical. And this is why I came here for help, it knew that the ratio was too simple and I needed more

17. anonymous

@Photon336 I'm going to tag you so that you know I need a little bit of viewing on this c:

18. anonymous

@Zale101 could you verify this? Determine the percent yield of this reaction, showing all steps of your calculation. % yield = actual yield / theoretical yield x 100 / 1 0.0028 /0.0014 = 2 * 100/1 = 200%

19. anonymous