1) the total ionic equation for the reaction BaCl2(aq) + Na2C2O4(aq) -----------> BaC2O4(s) + 2NaCl(aq)
Ba2+(aq) + 2Cl-(aq) + 2Na+(aq) + C2O42-(aq) ----------> BaC2O4(s) + 2Cl-(aq) + 2Na+(aq)
2) Net ionic equation for the reaction
Ba2+(aq) + C2O42-(aq) ----------> BaC2O4(s)
3) Moles of BaCl2 = molarity * volume in liters
= 0.0825 mol/L * 0.011 L
= 0.0009075 mol
Moles of NaC2O4 = molarity * volume in liters
= 0.0935 mol/L * 0.01 L
= 0.000935 mol
Hence BaCl2 is the limiting reactant
∴ The number of moles of BaC2O4(s) formed = moles of BaCl2 reacted
= 0.0009075 mol * 225.327 g/mol
= 0.2045 g
the theoretical yield in moles of barium oxalate (BaC2O4) formed = 0.0009075 mol
4) The molar mass of BaC2O4(s) = 137.327 g/mol + 2*12.0 g/mol + 4*16.0 g/mol
= 225.327 g/mol
5) The number of moles of BaC2O4(s) isolated = 0.1989 g / 225.327 g/mol
= 0.000883 mol
6) the percent yield for the student's experiment = (0.1989 g /0.2045 g)*100
= 97.26%