In: Chemistry
Sodium trioxalatocobaltate (III) trihydrate is
prepared by the following reactions:
[Co(H2O)6Cl2] + K2C2O4 *H2O >>> Co2O4 + 7 H2O
2CoC2O4 + H2O2 + 4 Na2C2O4 >>> 2 Na3[Co (C2O4)3*H2O + 2NaOH
what is the percent yield of Na3[Co (C2O4)3*H2O if 7.6 grams are obtained from 12.5 grams of [Co(H2O)6]Cl2 ? ( assume excess amount of the other reagents.)
[Co(H2O)6Cl2] + K2C2O4 *H2O -------------> Co2C2O4 + 7 H2O ) x 2
2 [Co(H2O)6Cl2] + 2 K2C2O4 H2O -------------> 2 Co2C2O4 + 7 H2O
2CoC2O4 + H2O2 + 4 Na2C2O4 ------> 2 Na3[Co (C2O4)3*H2O + 2NaOH
--------------------------------------------------------------------------------------------------------
2 [Co(H2O)6Cl2] + 2 K2C2O4 H2O + 4 Na2C2O4 -------------> 2 Na3[Co (C2O4)3*H2O + 2NaOH
475.8 g 807.8 g
12.5 g ?
mass of Na3[Co (C2O4)3*H2O = 12.5 x 807.8 / 475.8 = 21.22 g
thereotical yield = 21.22 g
actual yield = 7.6 g
percent yield = 7.6 x 100 / 21.22
= 35.8 %
percent yield = 35.8 %