Question

Write the balanced equation for the reaction between sodium and
water. Calculate the volume, mass and number of molecules of
hydrogen liberated when 460 gm. of sodium reacts with excess
of water at STP ?​

Answer 1

Answer:

Explanation:

2Na(g)+2H2O(l)------------->2NaOH(aq)+H2(g)

46g+36g------------------>80g+2g

As per the balanced equation

46 g of of Na gives 2 g of hydrogen.

460 g of Na gives------ ? g of hydrogen.

460x 2g / 46 g = 20g of Hydrogen.

1 gram molar mass of any gas at STP i.e., standard temperature 273 K and standard pressure 1 bar, occupies 22.4 litres known as gram molar volume.

2.0 g of hydrogen occupies 22.4 litres at STP.

20.0 g of hydrogen occupies------ ? litres at STP.

20.0 g x 22.4 litres / 2.0 g = 224 litres

2 g of hydrogen i.e., 1 mole of H2 contains 6.02 x {{10}^{23}} (H molecules 10 g of hydrogen contain =20*.02*10^23/2=6.02*10^24molecules

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