3.2 g of oxygen reacting with 3.2 g of mg to form magnicium oxide find the reactent wich is excess and by how much
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Mg will be excess of 2.133 gm
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Answered by
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2Mg + O₂ → 2MgO
2 mol 1 mol 2 mol
1 mol 1/2 mol 1 mol
3.2g of oxygen = 3.2/32 = 0.1 mol
3.2g of Mg = 3.2/24 = 0.133 mol
For 0.13 mol Mg oxygen should be 0.0665 mol, but we have 0.1 mol of oxygen. Therefore oxygen is excess. It is excess by 0.1 - 0.0665 = 0.0335 mol
Mass of 0.035 mol of oxygen = 0.0335*32 = 1.072g
Oxygen is excess by 1.072g
2 mol 1 mol 2 mol
1 mol 1/2 mol 1 mol
3.2g of oxygen = 3.2/32 = 0.1 mol
3.2g of Mg = 3.2/24 = 0.133 mol
For 0.13 mol Mg oxygen should be 0.0665 mol, but we have 0.1 mol of oxygen. Therefore oxygen is excess. It is excess by 0.1 - 0.0665 = 0.0335 mol
Mass of 0.035 mol of oxygen = 0.0335*32 = 1.072g
Oxygen is excess by 1.072g
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