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 mol-1 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 mol-1 510 x 100 = 51 000 g = 51 kg.