Question

calculate the volume mass and number of molecules of hydrogen liberated when 115gm of sodium reacts with excess of water at stp?

Answer 1

Answer:

30.115X

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