How many mL of a 0.1 M HCl are required to react completely with a 1gm mixture of Na2CO3 and NaHCO3 containing the equimolar amount of the two.
solution:let, the amount of Na2Co3 in the mixture = x gmNaHCo3 present in mixture = (1-x) gmMolar mass of Na2Co3 = 2*23+12+3*16 = 106 gm/molesimilarly, molar mass of NaHCo3 = 84 gm/moletherefore, moles of Na2Co3 in x gm = x/106 and moles of NaHCo3 in (1-x) gm = (1-x)/84According to the question,x/106 = (1-x)/84 so, x = 0.558gmThus the number of moles of Na2Co3 and NaHCo3 are 0.00526 moleAgain by reaction,
Na2CO3+2HCl2NaCl+H2O+CO2
NaHCO3+HClNaCl+H2O+CO21 mole ofNa2CO3
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Na2CO3+2HCl2NaCl+H2O+CO2
NaHCO3+HClNaCl+H2O+CO21 mole ofNa2CO3
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