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 0.346 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 8.84 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 mole of HCl remains after first step = number of mole of NaOH used
= (molarity of NaOH)*(volume of NaOH in L)
= (0.104*0.00884) mole
= (9.2*10^-4) mole

number of mole of HCl present intially = (molarity of HCl)*(volume of HCl)
= (0.0965*0.05) mole
= (48.2*10^-4) mole

number of mole oof HCl reacted in first step = (48.2*10^-4)-(9.2*10^-4)
= (3.9*10^-3) mole

balanced first step reaction is
Y2(CO3)3(aq) + 6HCl(aq) --> 2YCl3(aq) + 3CO2(g) + 3H2O(l)
6 mole of HCl react with 1 mole of Y2(CO3)3
1 mole of HCl react with 1/6 mole of Y2(CO3)3
(3.9*10^-3) mole of HCl react with (1/6)*(3.9*10^-3) mole of Y2(CO3)3
number of mole of Y2(CO3)3 = 0.65*10^-3

molar mass of Y2(CO3)3= 357.8 g/mol
mass of Y2(CO3)3 = (number of mole)*(molar mass)
= 0.65*10^-3*357.8
= 0.23 g

mass% of Y2(CO3)3 = {(mass of Y2(CO3)3)/(mass of sample)}*100
= (0.23/0.346)*100
= 66.5 %

Answer : 66.5%