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Question

What is the volume of 0.1 M HCl required to react completely with 1 g mixture of equimolar amount of Na2CO3 and NaHCO3?

A
106 mL
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B
128 mL
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C
156 mL
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D
212 mL
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Solution

The correct option is C 156 mL
Na2CO3nf=2+2HCl2NaCl+Na2CO3
NaHCO3nf=1+HClNaCl+CO2+H2O

Let x g of Na2CO3 and (1x) g of NaHCO3 are present in the mixture.

Equivalents of Na2CO3+ Equivalents of NaHCO3= Equivalents of HCl
x106×2+1x86×1=0.1×V ...(1)

Given, Moles of Na2CO3=Moles of NaHCO3
x106=1x86 ...(2)
86x=106106x
x=0.552

From equation (1),
0.552106×2+10.55286×1=0.1×V
0.0104+0.0052=0.1×V
0.0156=0.1×V
V=0.156 L
or, V=156 mL

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