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In an experiment, 4 gm of $ M_2O_x $ oxide was reduced to 2.8 gm of the metal. If the atomic mass of the metal is 56 gm/mol, the number of oxygen atoms in the oxide is (AFMC 2010)
$ 1 Mol M_2O_x = (2 \times 56 + 16x) gm$ $ Now, (2 \times 56 + 16x) gm of oxide = 112 gm metal $ $ \therefore 4 gm of oxide = { 112 \times 4 \over 112 + 16x } gm metal $ $ But { 112 \times 4 \over 112 + 16 x } = 2.8 (given ) \therefore x = 3 $
100 mL of $PH_3$ on heating forms P and $H_2$ . The volume change in the reaction is
$ 2PH_{3_{(g)}} \rightarrow 2P_{(s)} + 3H_{2_{(g)}} $ 2mL 3mL $ \therefore 100ml (?) $ = { 100 \times 3 \over 2 } = 150ml $ $ \therefore Increase = 50mL $
An ore contains 1.24% of the mineral argentite $ Ag_2S$ by mass. How many grams of this ore would have to be processed in order to obtain 1.0 g of pure solid silver ?
$ Ag_2S $ Ag ore $ Ag_2 S $ 248g 216g 100g 1.24g (?) $ \therefore1.0g (?) 1.148g = 1.148g $ = 92.58 \approx 92.6 g $
27 g of Al (at mass = 27) will react with oxygen equal to (IIT 1978)