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