Okay, so Cr is undergoing reduction right, as it goes from Cr5+ to Cr3+
Fe must therefore undergo oxidation. And it can only go from +2 to +3 state, so initially it must be Iron (II) thus option C is eliminated.
Then find mol Cr2O7 = cv = 0.100 x 0.01278 = 0.001278 mol
To find mol of Fe, you do the half equation:
Fe2+ ---> Fe3+ + e-
Then you multiply by 6, so the electrons cancel out in the full equation.
Ratio Cr2O72- : Fe2+ is 1:6
Therefore, multiply mol by 6.
n(Fe2+)= 0.001278 x 6 = 0.007668 mol
m(Fe2+)= 0.007668 x 55.9 = 0.43 g
Now you test out all the alternatives.
A - Mol Cl is twice the mol of Fe, so 0.007668 x 2 x 35.5 = 0.54
0.54 + 0.43 = 0.97 g - wrong
B - Mol SO4 is the same as the mol of Fe, so 0.007668 x (32.1 +64) = 0.74
0.74 + 0.43 = 1.17 g - wrong
C - we eliminated before
D - Mol NO3 is twice the mol of Fe, so 0.007668 x 2 x (14 + 48) = 0.95
0.95 + 0.43 = 1.38 g - RIGHT!
Absolute pain in the ass question for only one mark...