Answer to Question #101114 in Chemistry for Uzoma

Question

If 35.5g of ion was made to react with chlorine the mass of the product obtained would be either 80.1g or 103.01g depending on the amount of chlorine present.Which gravimetric ion will be in accordance?

Answer 1

Answer: From the given data we have to find out the mole of metal ion present in the given then we can find the actual metal ion name.

Depending upon the amount of Cl- ion present metal and can react which to define oxidation state and give two different products. Which can be 80.1 gram or 10 3.01 gram. This indicates metal ion present in two different oxidation State.

Consider x mol of metal ion present in 35.5 gram.

Hence, x = 35.5/M, M=Molecular weight.

If consider metal ion form the bi chloride salt.

Then, x mol MCl2 = 80.1 gm

So, 2x mol Cl- + x mol M2+ = 80.1

2x mol Cl- = 80.1 - 35.5 = 44.6.

[as x mol M2+ ion weight is 35.5 gm]

x = 44.6/2*35.4 = 0.629 mol.

So 1 mol of metal ion weight is = 56.3 gm/mol.

This molecular weight roughly matches with Fe2+ ion.

Hence iron is present as metal.

Now if higher amount Cl- ion present it can form FeCl3 also. So it can form only x mol( 0.638 mol) FeCl3.

x mol Fe2+ ion + 3x mol Cl- ion = 35.5 + 3*0.638*35.4 = 35.5 + 67.75 = 103.2 gm.(Given weight is 103.01 gm)

As proved by back calculation of weight of produced FeCl3 and estimation of molecular weight of unknown metal ion, which is found to be Fe.

Metal ion present is iron, first it from FeCl2 and second time FeCl3.