Question

calculate the volume ,mass and number of molecules of hydrogen liberated when 460g sodium react with excess of water at STP (atomic masses of Na=23U,O=16U,H=1U​

Answer 1

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=5x6.023x10^23
                                     =30.115X10^23

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