Calculate the volume mass and number of molecules of hydrogen when 230 g of sodium reacts with excess of water at stp
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Explanation:
2Na+2H₂O⁻⁻⁻⁻2NaOH+H₂
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
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