2.6 How many mL of are required to react completely with mixture of and containing equimolar amounts of both?
Total amount of mixture of Na2CO3 and NaHCO3 = 1 g.
Let the amount of Na2CO3 be x g.
So the amount of NaHCO3 will be equal to (1 - x) g.
Now it is given that it is an equimolar mixture.
So, Moles of Na2CO3 = Moles of NaHCO3.
or x = 0.558 g
It is clear that for 1 mol of Na2CO3 2 mol of HCl is required, similarly for 1 mol of NaHCO3 1 mol of HCl is required.
So number of moles required of HCl = 2(0.00526) + 0.0053 = 0.01578 mol
It is given that molarity of HCl is 0.1 which means 0.1 mol of HCl in 1l of solution.
Thus required volume :
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