Question

CoCl2.6H2O + H2C2O4.2H2O -> CoC2O4.4H2O + 2 HCl + 4 H2O

If, according to previously provided equation, the reaction uses 2.00 g of cobalt (II) chloride hexa hydrate and 1.26 g of  oxalic acid dihydrate, and the amount of cobalt oxalate tetrahydrate product actually collected is determined according to the following information:

* Mass of watch glass: 50.73 g

* Mass of filter paper: 0.38 g

* Mass of watch glass, filter paper and product: (5.22x10^1) g,

what is the percent yield of CoC2O4.4H2O? This time you have to determine the limiting reagent and the theoretical yield before determining the percent yield.

Answer 1

atomic weights : CO= 59, Cl= 35.5, H= 1, O=16

molar masses : CoCl2.6H2O= 59+71+6*18= 238 g/mole, H2C2O4.2H2O=2*1+2*12+4*16+2*18= 126 g/mole

CoC2O4. 4H2O= 59+24+64+4*18= 219 g/mole, moles of CoCl2.6H2O= mass/molar mass = 2/238= 0.008

moles of oxalic acid = 1.26/126= 0.01

theoretical moalr ratio of CoCl2.6H2O: Oxalic acid = 1: 1

actual ratio = 0.008 :0.01, the limiting reactant is COCl2.6H2O

hence moles of CoC2H4.4H2O =0.008 , mass of CO2C2H4.4H2O ( theoretically) =0.008*219 gm =1.752 gm

actual yield = mass of fliter paper, watch glass+ product- mass of filter-mass of watch glass= 52.2-50.73-0.38 =1.09 gm

% yield= 100*actual yield/theoretical yield= 100*1.09/1.752= 62.21%