## G5 from fundamentals!

manasa933
Posts: 72
Joined: Fri Sep 29, 2017 7:04 am

### G5 from fundamentals!

So I used the formula m(initial) x v(initial) = m(final) x v(final) and substituted the following:

m(initial) = 0.079 M
v(initial) = 250 ml
m(final) = 2.15 x 10 (to the power -3)

But in the solutions manual they didn't have the v(initial) anywhere.

Alyssa Pelak 1J
Posts: 72
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

### Re: G5 from fundamentals!

I used a stoichiometric approach because the mmol given are Na+ whereas the g and mL given are sodium carbonate. Therefore, you are supposed to find L of Na+
I notice you already found the molarity of the sodium carbonate (Na2CO3) to be .079M. I then took the 2.15 times 10^-3 mol Na+ and multiplied it by the molar ratio 2 mol of Na+ for every 1 mol Na2CO3. Then I multiplied it by 1L there is .079 mol of Na2CO3 (Molarity) to get an answer in mL.

venning-1J
Posts: 12
Joined: Fri Sep 29, 2017 7:04 am

### Re: G5 from fundamentals!

for part b,(b) 4.98 mmol CO3 ; would there be 3 moles of CO3 for every mole of Na2CO3?

Maria1E
Posts: 61
Joined: Sat Jul 22, 2017 3:01 am

### Re: G5 from fundamentals!

The ratio of CO3 to Na2CO3 is 1:1. The subscript 3 only applies to the oxygen, so there are 3 moles of oxygen for every 1 mole of NaCO3. However, the subscript does not apply to CO3 as a whole so there is only 1 mole of CO3 for every mole of Na2CO3.