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6.2g of phosphorus was reacted with excess oxygen to form phosphorus( V) oxide. Determine the mass of the oxide formed. (0 = 16.0; P 31.0)

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4P(s) + 5O2(g) → 2P2O5(g)
Moles of P =  6.2/31  = 0.2                           
Moles of P2O5 = ½ × 0.2 = 0.1moles
R.F.M of P2O5 = (31× 2) + ( 16×5) =142
Mass of P2O5 = 142 × 0.1 = 14.2g

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