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21 February, 13:26

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

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  1. 21 February, 17:24
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    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).
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