Question

If magnesium was the limiting reactant, calculate the theoretical yield of the gaseous product.

Answer 1

Mass of Magnesium strip (grams)
0.034 grams
Volume of gas collected
32 mL
Barometric pressure (atm)
1.10 atm
Room Temperature (ᐤC)
24 ᐤC
Vapor pressure of the water (torr)
22.4


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

Answer 2

What exactly is the reaction ! ?

Answer 3

I'm not sure Dx

I just have that data and 2 other questions before this one:

Write the balanced equation for the reaction conducted in this lab, including appropriate phase symbols.
Mg(s) + 2 HCl(aq) -> H2(g) + MgCl2(aq)

Determine the partial pressure of the hydrogen gas collected in the gas collection tube.
22.4/760 = 0.02947 atm
1.1 - 0.02947 = 1.07 atm

Answer 4

okai now this makes sense !

Answer 5

You've got me, I can't understand chemistry if my life depended on it.

So what if magnesium were the limiting reactant?

Answer 6

looking at the reaction we can see that H2 produced and Mg in reactant side are having the same stoicheometric number

Answer 7

do you get that part?

Answer 8

I kinda do, just not on a true level of understanding.

Answer 9

Do you understand what the limiting reagent does?

Answer 10

hey @taramgrant0543664 U carry out then !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 11

@Lucker do you understand what the limiting reagent does? If you do awesome, if you don't then I'll let you know since it's important to understand for the question

Answer 12

I understand that with a limiting reactant, it limits the amount of material used in the reaction.
So if you were to have a certain amount of the reactant to a ratio of another, the limiting reactant will restrict how much of the other will be used.

Answer 13

Right so since we are saying Mg is our limiting the HCl is in excess for the reaction

Answer 14

Mg(s) + 2 HCl(aq) -> H2(g) + MgCl2(aq)
We know that Mg and H2 are in a 1:1 ratio due to the stoiciometric coefficients

Answer 15

exactly, so if Mg were to be the limiting reactant, what would the theoretical yield be? If it's 1:1, we won't have to do anything like consider for every 2 of Mg will be 1 of HCl, it's just a straight forward 1:1

Answer 16

Yes so if we figure out the moles of Mg that we have that number of miles will be the same as the amount of H2 that will be in the products due to the 1:1 ratio
To find the moles we have the mass of 0.034g of Mg so we divide by the molar mass of Mg
n=0.034/24.03

Answer 17

so the moles are 0.0014(1489)

I'm not sure what to find of HCl though, just the molar mass (34.46)

Answer 18

Yes the moles are 0.0014 moles of H2 to determine the theoretical yield is normally in mass so you would multiply the 0.0014 by the molar mass of H2 (so 2.02g/mol)
Are you required to know how much HCl is used?

Answer 19

Ah, that was the problem, I was focusing on HCI

Answer 20

You can determine how much HCl was used if you want we know that Mg is 0.0014moles
Mg and HCl are in a ratio of 1:2
0.0014 moles Mg x (2 moles HCl/1 mole Mg) x (34.46g/mol / 1mol HCl)
0.0014x 2x 34.46= amount of HCl used in the reaction

Answer 21

No no, you were right about H2, my problem was that I thought that I was looking for HCl instead >.<

Answer 22

Hana that's ok! Little mistakes happen to everyone, happy that we got that cleared up though!!

Answer 23

alright so 0.0014 * 2.02 is 0.0028, that's it right, just multiplying them? I hated doing yields, I always feel like there's more to them

Answer 24

Yes the 0.0028g is the theoretical yield for H2
It does seem a little short but that's all you have to do for it

Answer 25

alright, thank you Tara c:

Answer 26

No problem! Happy to help!