Question

Question1: you begin with 2.0g of Fe(NH4)2(SO4)2*6H2O. How many moles is this? 5.1e-3
Question 2: you then add 10mL of 1.0 M oxalic acid, H2C2O4. How many moles is this?

Answer 1

I was able to find the answer to #1, (5.1e-3). But I'm not sure what to do with question two? Im guessing I convert the mL to g and then continue by diving by molar mass, & then add those moles to the ones found in question 1?

Answer 2

use the molarity equation for #2: \(Molarity=\dfrac{moles_{solute}}{L_{solution}}\)

note that you have to use liters, not mL.

Answer 3

ahhh! ok ! and for the moles of solute would i use the value that i found in question 1?

Answer 4

hm the moles of solute is the value you are finding since you're given molarity and volume.
Im not sure if they want you to add both values (moles from #1), but thats how i would interpret it.

Answer 5

hm ok so I would set up my equation as (0.01)*(5.1e-3)=moles?

Answer 6

remember you're using the L of solution and the molarity (not the moles previously found), so moles = 0.010 L *1.0 M

Answer 7

so assuming no reactions occur, just add the moles from #1 and the moles from #2

Answer 8

OH! gosh i over looked the "1.0 M"

Answer 9

lol it happens

Answer 10

haha yes! hm that seemed so simple!

Answer 11

lol because it is. chemistry is straight forward, most of the time