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Calculate the volume ,mass and number of molecules of H2 liberated when 230g of sodium reacts with excess of water at STP.

(at.mass of Na=23U,O=16U,H=1U


so 46 g of Na reacts with water to give 22.4 L of H₂
so 230g of Na reacts with water to give 22.4X5=112 L of H₂
112 L of H₂=112/22.4=5 moles of H₂
mass of H₂ released=5X2=10g
number of molecules=5x6.023x
                                     =30.115X 10^{23}
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