How many grams of iron (iii) chloride must decompose to produce 78.4 milliliters of iron metal, if the density of iron is 7.87 g/ml? show all steps of your calculation as well as the final answer. fecl3 → fe + cl2

First step in any case of chemistry BALANCE THE EQUATION 
so you have 2FeCL3 --> 2Fe + Cl2 

Ok second step start with the number with only one unit, not the one comparing two units 

So you have 78.4 mL of Fe (III) 
so mulitple the milliters by the conversion of g/mL to get grams 
78.4 mL Fe(III) * (7.87 g / 1 mL) = 617 grams of Fe (III) 

Next convert grams to moles since moles is the only unit that can be used to compare amounts of different substances equally. To do this divide by the Molar Mass of Iron ... or multiply by the reciprocal 

617 g Fe(III) * ( 1 mol / 55.85 grams ) = 11.1 moles of Fe 

Next use the moles to calculate the moles of Iron Chloride. From our balanced equation is a 2:2 mole ratio 

So 11.1 moles of Fe * (2 mol FeCl3/ 2 mole Fe) = 11.1 mole FeCl3 

Next use the Molar Mass of FeCl3 to convert the mole of FeCl3 back into grams for your final answer 

11.1 mole FeCl3 * (162.2 g FeCl3/ 1 mole FeCl3) = 1791.89 grams 

Remember Significat Figures count so round your answer to 3 digits 

Giving you 1790 grams as your final answer .. or 1.79 kg 

hope it helps