# Calculations 2

by Janka
Tags: calculations
 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 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.
Mentor
P: 8,272
 Quote by Janka 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 :(
You should, somewhere, have a formula: molarity=(number of moles) /(volume of solution). This will be useful here.

 (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. here i dunno what to do as well...:(
Well, i would calculate the number of moles in 50kg of H2SO4, use the equation to see how many moles of CaCO3 are required to neutralise it, then find the mass of this number of moles of CaCO3.
 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.
HW Helper
PF Gold
P: 3,725

## Calculations 2

When doing these calculations, please show your units. They will be a great help to you. Make sure that the units for concentration and volume of solution are similar.

What is the '1000' doing in your calculation? Units?

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