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
