DentistWannaBe Posted September 6, 2006 Report Share Posted September 6, 2006 6 volt lead storage battery contain 700 g of pure H2SO4 (l) dissolved in water. the battery is accidently dropped, spilling its contents. To neutralize the acid, solid sodium carbonate, Na2CO3 is used. Harmless Na2SO4 is formed and CO2 gas is released. If a 2.0 M solution of sodium carbonate was used to neutralize the spilled acid, how many liter are need? I got stucked on this question, can you solve this question. If so, plz show ur calculation. Thank you!! Link to comment Share on other sites More sharing options...
bahamabreeze Posted September 7, 2006 Report Share Posted September 7, 2006 You'll need around 3.5L. Molecular mass of H2SO4 is around 98 (let's say 100 to make things easy, you can calculate exactly how much by yourself), so with 700g you get 7 moles. Since it's basically 1:1 reaction, you'll need 1 mole of Na2CO3 to neutralize 1 mole of H2SO4, so you need 7 moles. If your concentration is 2M, then you'll be needing 3.5L (7 moles/2 moles/L = 3.5 L) H2SO4+Na2CO3 -- Na2SO4 + CO2 + H20 Bahamabreeze Link to comment Share on other sites More sharing options...
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