Question

calculate the volume, mass and no. of hydrogen liberated when 230 gr of sodium reacts with excess water at STD ​

Answer 1

Answer:

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}10

23

=30.115X10^{23}10

23

Explanation:

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