If the decomposition of 3.785 g iron (III) chloride yields 1.302 g iron and 2.483 g chlorine, how many grams of iron and chlorine are produced when 2.175 g iron (III) chloride decomposes?

See my response to Ray.

http://www.jiskha.com/display.cgi?id=1285730192

To find the grams of iron and chlorine produced when 2.175 g of iron (III) chloride decomposes, we can use the concept of stoichiometry.

First, let's find the molar mass of each compound involved:
- Molar mass of Iron (III) Chloride (FeCl3) = (1 x molar mass of Iron) + (3 x molar mass of Chlorine)
= (1 x 55.845 g/mol) + (3 x 35.453 g/mol)
= 162.205 g/mol

Now, let's calculate the number of moles of Iron (III) Chloride in 2.175 g:
- Number of moles of Iron (III) Chloride = Mass / Molar mass
= 2.175 g / 162.205 g/mol
≈ 0.0134 mol

According to the balanced chemical equation, 1 mole of Iron (III) Chloride decomposes to produce 1 mole of Iron and 3 moles of Chlorine.

So, the number of moles of Iron produced = 0.0134 mol
And the number of moles of Chlorine produced = 3 x 0.0134 mol = 0.0401 mol

Finally, let's calculate the grams of Iron and Chlorine produced:
- Mass of Iron produced = Moles of Iron produced x Molar mass of Iron
= 0.0134 mol x 55.845 g/mol
≈ 0.747 g

- Mass of Chlorine produced = Moles of Chlorine produced x Molar mass of Chlorine
= 0.0401 mol x 35.453 g/mol
≈ 1.421 g

Therefore, when 2.175 g of Iron (III) Chloride decomposes, approximately 0.747 g of Iron and 1.421 g of Chlorine are produced.