Sodium carbonate nuetralizes H2SO4 as shown below in the reaction:
Na2CO3 (s) + H2SO4 (aq) -->Na2SO4 (aq) + H2O (l) +CO2 (g)
How many grams of sodium carbonate are required to nuetralize 455ml of 5.4% ( W/V) solution of H2SO4?
Calculate mass of solute in gm
mass of solute = concentration of solute ( w/v %) volume of solution in ml / 100
mass of H2SO4 = 5.4 455 / 100 = 24.57 gm
molar mass of H2SO4 = 98.079 gm / mole
then 24.57 gm of H2SO4 = 24.57 / 98.079 = 0.2505 mole of H2SO4
According to reaction H2SO4 react with Na2CO3 in equimolar proportion therefore to react with 0.2505 mole of H2SO4 required Na2CO3 = 0.2505 mole
molar mass of Na2CO3 = 105.9888 gm /mole then 0.2505 mole of Na2CO3 = 0.2505 105.9888 = 26.55 gm
26.55 gm of Na2CO3 required for complete nertrilization
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