Question

1.15g of a metallic element reacts with 300cm3
of oxygen at 298K and 1atm pressure, to form
an oxide which contains O2– ions.
What could be the identity of the metal?

1)Calcium
2) Strontium
3) Barium
4) Sodium

Answer 1

@UnkleRhaukus @Preetha @ParthKohli @satellite73

Answer 2

can you find the number of moles of Oxygen?

Answer 3

\[pV=nRT\]

Answer 4

you'll have to be careful of units

Answer 5

yes 0.012 mole

Answer 6

how did you get that?

Answer 7

sorry moles of O2 = 300/24000 = 0.00125 mol

Answer 8

tht is 0.012 mole

Answer 9

yeah sorry. that is right 0.0125 mol of O_2

Answer 10

\[\text M+\text O_2\to \text M_2\text O\]

Answer 11

can you balance that equation?

Answer 12

yes
4M + o2 ---> 2M20

Answer 13

so how many mole of the Metal reacted?

Answer 14

how do i find tht?

Answer 15

4 * 0.0125

Answer 16

right

Answer 17

thanks alot very simple question, but why would they do include pv = nrt

Answer 18

M+O2→M2O
how wuld u know this?

Answer 19

i was just guessing, i havent studied chemistry in a while, i still think pv=nrt would have worked,, but it wasent the quickest way

Answer 20

i got that from the fact that the ions were O2– ions

Answer 21

the is an even simpler way , because only one of the four options will bond with O^-2 ions

Answer 22

have you got to the answer yet?

Answer 23

i dont understand that point

Answer 24

yes its NA

Answer 25

yeah

Answer 26

actually i dont know if what i just said makes any sense at all ,
\[\cancel{\cancel{\text{because only one of the four options will bond with O^-2 ions }}}\]scratch that

Answer 27

wait what about Ca,

Answer 28

M + O_2 → M^a + O^2- → M_2O_a

Answer 29

now i dont know how i got that formula . hmm . sorry for all the confusion.

Answer 30

hopefully someone who knows what there doing will be able to help you better, i have forgotten my chem. ):