Question
Sat June 16, 2012 By: Priyanka Kumar

How many 0.1M HCl are required to react completely with 1g mixture of sodium carbonate( na2co3) and sodium hydrogen carbonate(NaHCO3) containing equimolar amounts of both.

Expert Reply
Wed June 20, 2012
Let the mass of Na2CO3 = x
mass of NaHCO3 = (1-x)
Because of equal molarity
x/106  = 1-x/84
x= 0.558g  , 1-x = .442g
By equation Na2CO3 + HCl  ?2NaCl + H2O + CO2
106 g of Na2CO3 require HCl = 73g so .558 Na2CO3 require HCl = 73/.558 /106 = 0.384g
similarly  NaHCO3+ HCl  ?NaCl + H2O + CO2
 
0.442g NaHCO3 require HCL = 36.5 x .442 / 84 = 0.192g
so total mass of HCl = 0.384 +0.192 = 0.576g
Volume of solution= 0.576 X 1000/ 36.5 X 0.1n = 0.1578 L or 157.8 mL  (molarity = mass of HCl / molar mass of HCl / volume of solution,.
 
 
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