anonymous
  • anonymous
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
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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: \(\mathsf{Number} \space \mathsf{of } \space \mathsf{moles } = \cfrac{\textbf{Given mass}}{\textbf{Molar Mass}} \) Molar mass of 2 Nitrate ions (anions) = \(\mathsf{2*(14+48)} \) = \(\mathsf{124 gm}\) . And the Atomic mass of oxygen is 16 gm. So the oxide has weight of : \(\mathsf{5-3.29 g = 1.71 g}\) . Now as metal oxide and metal nitrate have equal no. of moles, so : \(\cfrac{\textbf{Given mass of metal nitrate}}{\textbf{Molar mass of metal nitrate}} = \cfrac{\textbf{Given mass of metal oxide}}{\textbf{Molar mass of metal oxide}} \) 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 : \(\cfrac{5}{x+124} = \cfrac{1.71}{x+16} \) \(5(x+16) = 1.71(x+124) \) \(5x + 80 = 1.71 x + 212.04 \) \(5x - 1.71 x = 212.04 - 80 \) \(3.29 x = 132.04\) \(x = 40 (\mathsf{Approx.} )\) 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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