Question

calculate the mass and volume of H2 gas released when 460 gm of sodium reacts with excess of water at STP ? [Na23U,O-16U, H-1U]

Answer 1

Answer: Mass = 20g, Volume = 224litres

Explanation:

460g of Na = 20 moles.

2Na+2H2O = 2NaOH+H2

So from this balanced equation, we can understand that for every 2 moles of Na 1 mole of H2 is released. So for 20 moles of Na, 10 moles of H2 will be released.

1 mole of H2 weighs 2g. Then 10 moles will weigh 20g.

According to Avagadro's law, one mole of any gas will occupy a volume of 22.4l at STP.

Then 10 moles will occupy 22.4*10 l volume = 224l.

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