A)Â Â Â mass percent of iron in the ore = mass of
iron/ mass of iron ore x 100
                                                       Â
= 242 grams / 500 grams x 100
                                                        Â
= 48.4 %
B)Â Â 2 Na3PO4 +
3Ba(NO3)2 -------------> 6NaNO3
+ Ba3(PO4)2
C)
    mass of iron is present in a 6.03 gram
sample of iron (II) chloride
        = [molar mass
of Fe / molar mass of FeCl2] x mass of sample
        = [55.8 g /
126.7 g ] x 6.03 grams
        = 2.65
grams
