Question

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

Answer 1

300 cm^3 of oxygen is 300/24000= 0.0125 moles

Now you find the no. of moles of each of the elements in 1.15g
Ca= 1.15/40= 0.02875
Mg= 1.15/24= 0.0958
K= 1.15/39= 0.029
Na= 1.15/ 23= 0.05

The equations for the oxidation of these elements are as follows:
\[2Ca+O2\rightarrow2CaO\]
\[2Mg+O2\rightarrow2MgO\]
\[4Na+O2\rightarrow2Na2O\]
\[4K+O2\rightarrow2K2O\]

For the first two, the ratio is 2:1
For the last two, the ratio is 4:1

So now, with the no. of moles of oxygen and the metals that you have calculated you see which ones satisfy the above ratios.

Here, it is sodium metal. Since the no. of moles of oxygen are 0.0125. 4 times 0.0125= 0.05 which is the no. of moles of Na. So your answer is D.

Answer 2

thanks helped alot

Answer 3

yw.