Question

How many moles of magnesium chloride are formed when 55 mL of 0.70 M HCl is added to 42 mL of 1.204 g Mg(OH)2?

Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)

A. 0.021 mol
B. 0.019 mol
C. 0.012 mol
D. 0.039 mol

Answer 1

@aaronq Can you please help me with this question?

Answer 2

are you sure you copied the question right?

this "42 mL of 1.204 g Mg(OH)" doesnt make sense

Answer 3

Well it's supposed to be Mg(OH)2

Answer 4

i mean this part "42 mL of 1.204 g"

Answer 5

Yep it's right...

Answer 6

do you think they meant "42 mL of 1.204 M" or "added to 1.204 g Mg(OH)2"

you can't add 42 mL of a solid

Answer 7

yeah I think that's what they meant.

Answer 8

the second right?

so you would first find the limiting reagent.

to do this you convert both reactants to moles, then divide each by it's coefficient in the balanced equation. Whichever has less after this division is the limting reagent

Answer 9

Ok so the answer is C right?

Answer 10

idk, i havent worked it out. i can check your steps if you post them

Answer 11

0.038 x (1 mol MgCl2/2 mols HCl) = approx 0.038 x 1/2 = approx 0.019
0.05 x (1 mol MgCl2/1 mol Mg(OH)2) = 0.0 x 1/1 = approx 0.05

Answer 12

how did you get 0.05?

Answer 13

mols HCl = M x L = approx 0.038
mols Mg(OH)2 = approx 0.05

Answer 14

the moles of magnesiums hydroxide are 0.0206448 moles

Answer 15

oh oops :/

Answer 16

so you'd use the one with less moles.. which is HCl

Answer 17

then what?

Answer 18

then find the moles MgCl2 as you would normally

\(\sf \dfrac{moles~of~HCl}{HCl's ~coefficient}=\dfrac{moles~of~MgCl_2}{MgCl_2's ~coefficient}\)

Answer 19

plug in what you know then solve (for moles of MgCl2)

Answer 20

oh ok so it is C lol. thanks for your help :)

Answer 21

nope, it's B