Question

Part A: Synthesis of Metal Oxalate

Metal Salt (Circle):

FeCI2 * 4H2O

Ni(NO3)2 * 6H2O

MnSO4 * H2O

Metal: 4.079

Filter Paper: .178

Metal Oxalate Recovered: 3.071 g

Theoretical Yield: ??

Percent Yield: ??

Part B:

Calculate theoretical yield for each of the possible decomposition products, determine the identity of the product, and calculate the percent yield.

Aluminum boat weight: 1.871

Metal decomposition product recovered: 0.972 g

Theoretical yield of product of reaction A: ??

Theoretical yield of product of reaction B: ??

Theoretical yield of product of reaction C: ??

Theoretical yield of product of reaction D: ??

Decomposition Product Idenitity: ??

Answer 1

Part A.

i) The no. of moles of FeCl2.4H2O = 4.079 g/198.805 g.mol^-1 = 0.020518 mol

The theoretical yield of Fe(C2O4)2 = 0.020518 mol * 231.881 g/mol = 4.758 g

The percent yield = (3.071 g/4.758 g)*100 = 64.54%

ii) The no. of moles of NiCl2.6H2O = 4.079 g/237.683 g.mol^-1 = 0.017162 mol

The theoretical yield of Ni(C2O4)2 = 0.017162 mol * 234.729 g/mol = 4.028 g

The percent yield = (3.071 g/4.028 g)*100 = 76.24%

iii) The no. of moles of MnSO4.H2O = 4.079 g/169.009 g.mol^-1 = 0.024135 mol

The theoretical yield of Mn(C2O4)2 = 0.024135 mol * 230.974 g/mol = 5.574 g

The percent yield = (3.071 g/5.574 g)*100 = 55.09%