Question

The amount of hydrochloric acid added to the reaction was more than enough to consume all of the magnesium. Therefore, the magnesium is the limiting reactant. How many milliliters of 12 ­M hydrochloric acid solution necessary to consume 0.080 grams of magnesium metal?
How do I do this?

Answer 1

First we need the balanced reaction
Mg (s) + 2 HCl (aq) --> MgCl 2 (aq) + H2 (g)

Start with .08g of Mg. We need moles of Mg. So .08g/24g = .0033mole
which i guess makes sense considering the super tiny initial .08grams of Mg.

Answer 2

also to be brutally honest im not 100% sure i did this correctly

Answer 3

Thats what I had so far except I made Mg 24.3, I think after ".0066 moles of HCl" you have to convert it to liters, so wouldn't it be 1L/12molesHCl? rather than "12mole/liter"

Answer 4

than 1000mL/L

Answer 5

Yea, your correct. I actually did the same thing. In the problem i got to --> .0066moles of HCl * (1L/12moles HCl) which was where the .00055Liters came from.

Answer 6

Then i just multiplied by 1000.

Answer 7

yeah your right

Answer 8

with its .055ml or .55ml? because I got .55mL

Answer 9

0.080gMg X 1moleMg / 24.3g Mg x 2molesHCl / 1moleMg x 1L HCl / 12moleHCl x 1000mL / L = thats how I wrote it

Answer 10

wait*

Answer 11

one sec getting my calculator

Answer 12

Your right. I was rounding my numbers to 2 significant numbers at each step when I should have left that step for the end. I did the calculations again and I agree with you I got .55mL. Weird how rounding caused that big of an error though...

Answer 13

Sorry for the screw up. Its been a year since AP chem.

Answer 14

yeah SF suck, I thought u might have done a typo haha

Answer 15

its cool, I noticed

Answer 16

Yea... i can never keep SF straight. Seems like you got your material down solid though so nice work.

Answer 17

Thanks hopefully I can retain it all for years