A 5.00 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).