1
answer
0
watching
204
views
13 Dec 2019
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?
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?
Elin HesselLv2
17 Dec 2019