Calculate rhe mass of hydrogen liberated when 230 grams of sodium reacts with excess of water at STP
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Answered by
20
Atomic or molecular mass of Na = 23
230 gms = 10 moles
2 Na + 2 H2O ==> H2 (g)+ 2 Na OH
2 moles of Sodium give 2 grams or 1 mole of H2. So 10 moles give grams.
230 gms = 10 moles
2 Na + 2 H2O ==> H2 (g)+ 2 Na OH
2 moles of Sodium give 2 grams or 1 mole of H2. So 10 moles give grams.
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Answered by
37
see the reaction,
2Na + 2H₂O → 2NaOH + H₂
you can see that
2mole of Na form 1 mole hydrogen gas
46g of Na form 2g hydrogen gas
230g of Na form 2*230/46 g hydrogen gas hence
the amount of H₂ = 10 g
2Na + 2H₂O → 2NaOH + H₂
you can see that
2mole of Na form 1 mole hydrogen gas
46g of Na form 2g hydrogen gas
230g of Na form 2*230/46 g hydrogen gas hence
the amount of H₂ = 10 g
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