A5.00 g sample of an anhydrous group ii metal nitrate loses 3.29g in mass on strong heating. which metal is present? a magnesium b calcium c strontium d barium

When group II metal nitrate is heated, it produces metal oxide and nitrogen dioxide gas, and the ratio between metal nitrate and metal oxide is 1:1 .

From 1 mole nitrate you get 1 mole oxide.

As:
Number of moles = Given Mass/ Molar Mass

Molar mass of 2 Nitrate ions (anions) = 2*(14+48) = 124 gm.

And the Atomic mass of oxygen is 16 gm.

So the oxide has weight of : 5-3.29 g = 1.71 g

Now as metal oxide and metal nitrate have equal no. of moles, so :

Given mass of metal nitrate/Molar mass of metal nitrate = Given mass of metal oxide/Molar mass of metal oxide

Let us say that the mass of the metal is x grams.
So, molar mass of metal nitrate = x + 124 grams
Molar mass of metal oxide = x + 16 grams.
Therefore , we get : 5/x+124 = 1.71/x+16
5(x+16) = 1.71(x+124)
5x + 80 = 1.71 x + 212.04
5x - 1.71 x = 212.04 - 80
3.29 x = 132.04
x = 40 (Approx.)

Therefore, the mass of the metal used is 40 grams. Therefore it is Calcium as it has mass of 40 grams (atomic mass) . 

Because it was a dramatically big guy