Solid iron(III) hydroxide decomposes to produce iron(III) oxide and water vapor. If 0.75 L of water vapor is produced at STP, how many grams of iron(III) hydroxide were used?

Question

xishem · Answer

First, let's write and balance the chemical equation for the decomposition of iron(III) hydroxide into iron(III) oxide and water vapor...$$2Fe(OH)_3 \rightarrow Fe_2O_3+3H_2O$$ Now, let's find out how many moles of water are being produced...
$$PV=nRT \rightarrow n=\frac{PV}{RT}$$$$n=\frac{(0.986atm)(0.75L)}{(0.08206\frac{L*atm}{K*mol}(273.15K)}=0.03299mol\ H_2O$$And then let's convert from moles of water to moles of iron(III) hydroxide...$$0.03299mol\ H_2O*\frac{2mol\ Fe(OH)_3}{3mol\ H_2O}=0.022mol\ Fe(OH)_3$$And finally to moles of iron(III) hydroxide...$$0.021993mol\ Fe(OH)_3*\frac{106.88g\ Fe(OH)_3}{1mol\ Fe(OH)_3}=2.4g\ Fe(OH)_3$$