4g mixture of sodium carbonate and sodium chloride reacts completely with 50 mili equivalent of HCL. Calculate the percentage of sodium carbonate

4g mixture of sodium carbonate and sodium chloride reacts completely w

1.	4g mixture of sodium carbonate and sodium chloride reacts completely with 50 mili equivalent of HCL. Calculate the percentage of sodium carbonate
Answer» LET the mixture contains x g of sodium carbonate and 1−x g of sodium bicarbonate.The molar masses of sodium carbonate and sodium bicarbonate are 106 g/mol and 84 g/mol respectively The number of moles of sodium carbonate and sodium bicarbonates are 106x and 841−x respectively.Since, it is an equilmolar mixture,106x = 841−x 84x=106−106x190x=106x=0.5579Number of moles of sodium carbonate = 1060.5579 =0.005263Number of moles of sodium hydrogen carbonate = 841−0.5579 =0.005263One MOLE of sodium carbonate will react with 2 moles of HCL and 1 mole of sodium bicarbonate will react with 1 mole of HCl.Total number of moles of HCl that will completely neutralize the mixture =2×0.005263+0.005263=0.01578 molesVolume of 0.1 M HCl required = 0.10.01578 =0.158L=158 mL.Explanation:This may help you