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Calculations 2by Janka
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#1
May2307, 06:16 AM

P: 16

Hello people...could u help me pls...coz i am rly stuck in these thing like calculations...i know all right answers but i am not too sure how to get them...please if somebody of us has a few minutes could u help me with that?
Thank you very much. 50 kg of pure sulphuric acid were accidentally released into a lake when a storage vessel leaked.Two methods were proposed to neutralise it. (a) The first proposal was to add a solution of 5 M NaOH to the lake. Sodium hydroxide reacts with sulphuric acid as follows: 2NaOH (aq) + H2SO4(aq) > Na2SO4 (aq) + 2H2O (l) Calculate the volume of 5 M NaOH required to neutralise the sulphuric acid by answering the following questions. (i) How many moles of sulphuric acid are there in 50 kg of teh acid? H2SO4 = 2+32+ 16+16+16+16= 98 g mol1 50 kg = 50 000g / 98 = 510.2 moles (ii) how many moles of sodium hydroxide are required to neutralise this acid? 510 x 2 = 1020 moles (iii) Calculate the volume , in dm3 , of 5 M NaOH which contains this number of moles. well..here i dunno what to do :( (b) The second proposal was to add powdered calcium carbonate which reacts as follows: CaCO3 (s) + H2SO4 (aq) > CaSO4(s) + H2O(l) +CO2(g) Calculate the mass of calcium carbonate required to neutralise 50 kg of sulphuric acid. no.of moles in 50 kg of H2SO4 = 510 moles CaCO3 = 40+12+(16x3) =100 g mol1 510 x 100 = 51 000 g = 51 kg. 


#2
May2307, 06:37 AM

Mentor
P: 8,305




#3
May2307, 07:16 AM

P: 16

What you think about this calculation :
(iii) Calculate the volume , in dm3 , of 5 M NaOH which contains this number of moles. I used this formula: amount of substance, in mol= concentartion x ( volume of solution / 1000) 1020 = 5 M x ( V / 1000 ) 1 020 000 = 5 M x V 204 000cm3 = V 204 000 cm3= 204 dm 3 hope it s right now. 


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