a 563-mg sample containing only CaCO3 (100.1 mg/mmol) and SrCO3 (147.6 mg/mmol) is dissolved in an acid; Na2C2O4 is added to precipitate CaC2O4 (128.1 mg/mmol) and SrC2O4 (175.6 mg/mmol). The oxalate precipitate is filtered and dried to yield 703 mg of the anhydrous mixed oxalates. Calculate the percentages of CaCO3 and SrCO3 in the mixture. Show ALL your work!!
2007-10-04 05:25:21 · 3 answers · asked by macjack 1 in Science & Mathematics ➔ Chemistry
given X is mol of Ca2+
and Y is mol of Sr2+
a 563-mg sample containing only CaCO3 (100.1 mg/mmol) and SrCO3 (147.6 mg/mmol)
means 100.1X+147.6Y=563
The oxalate precipitate is filtered and dried to yield 703 mg of the anhydrous mixed oxalates
means 128.1X+175.6Y=703
When u determine the value of X and Y, u can reach weight of CaCO3 and SrCO3
mg CaCO3=100.1X
mg SrCO3=147.6Y
2007-10-04 05:52:16 · answer #1 · answered by Merci Pians 1 · 1⤊ 0⤋
CaCO3 + 2H+ -> Ca2+ + H2O + CO2
SrCO3 + 2H+ -> Sr2+ + H2O + CO2
All reactions are 1:1 ratio.
assume CaCO3 is x mg, SrCO3 is 563-x mg.
x / 100.1 = mmol of CaCO3 or Ca2+
563-x / 147.6 = mmol of SrCO3 or Sr2+
then because this mmol of Ca2+ will obtain the same amount of CaC2O4 in terms of mmol, same for Sr2+
Ca2+ + C2O42- -> CaC2O4
Sr2+ + C2O42- -> SrC2O4
x / 100.1 * 128.1 + 563-x / 147.6 * 175.6 = 703
CaCO3: x / 563 * 100%
SrCO3: 100% - CaCO3
2007-10-04 12:57:13 · answer #2 · answered by Carborane 6 · 0⤊ 0⤋
Write down the equations and then transfer everything to moles.
2007-10-04 12:48:45 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋