Question

Y₂(CO₃)₃(aq) + HCl(aq) --> YCl₃(aq) + CO₂(g) + H₂O(l)
HCl(aq) + NaOH --> NaCl(aq) + H₂O(l)


Consider the unbalanced equations above. A 0.283 g sample of impure yttrium carbonate was reacted with 50.0 mL of 0.0965 M HCl. The excess HCl from the first reaction required 5.31 mL of 0.104 M NaOH to neutralize it in the second reaction. What was the mass percentage of yttrium carbonate in the sample?

Answer 1

number of excess HCl = number mole of NaOH
= (molarity)*(volume in L)
= 0.104*0.00531
= (5.5*10^-4) mole

number of mole of HCl initially = 0.05*0.0965
= (48.2*10^-4) mole

number of mole of HCl reacted in first step= (48.2*10^-4) - (5.5*10^-4)  
= (42.7*10^-4) mole

balanced first step reaction is
Y2(CO3)3 + 6HCl -- > 2YCl3 + 3CO2 + 3H2O
number of Y2(CO3)3 = (number of mole of HCl used)/6
= (42.7*10^-4)/6
= (7.12*10^-4) mole

molar mass of Y2(CO3)3 = 297.9 g/mol
mass of Y2(CO3)3 = (number of mole)*(molar mass)
= (7.12*10^-4)*297.9
= 0.2121 g

mass% of Y2(CO3)3 = {(mass of Y2(CO3)3)/(total mass of sample)}*100
= (0.2121/0.283)*100
= 74.9%

Answer : 74.9 %