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

Question

Y2(CO3)3(aq) + HCl(aq) YCl3(aq) + CO2(g) + H2O(l) HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) Consider the unbalanced equations above. A 0.730 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 6.10 mL of 0.104 M NaOH to neutralize it in the second reaction. What was the mass percentage of yttrium carbonate in the sample?

Explanation / Answer

excess HCl = moles of NaOH = ( 6.1 x0.104/1000) = 0.0006344 moles , moles of HCl = 50 x0.0965/1000 = 0.004825 , moles of HCl reacted = 0.004825-0.0006344 = 0.00419 , Y2(CO3)3 + 6HCl ---> 2YCl3 +3 CO2 + 3H2O, moles of Y2(CO3)3 = ( 0.00419/6) = 0.0006984 moles , mass of Y2(CO3)3 = 0.0006984 x357.84 = 0.2499 gm mass % = ( 0.2499/0.73) x100 = 34.23 %

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