Chemistry Help? I don't know how to do this problem!

Iron(ii) hydroxide reacts with oxygen gas and water vapor to produce iron(iii) hydroxide. What mass of product can be produced by the complete reaction of 55 mL of oxygen gas at 1.8atm and 345 degrees Celsius with excess iron(ii) hydroxide and water vapor?

That's an 2 step problem.

Use the ideal gas law (PV=nRT) first to figure out how many mols of O2 you have.
PV=nRT, solve for n, mols O2
T= Kelvin, ALWAYS: K=C+273 = 618
1.8(0.055) = n(0.08206)(618)
0.099 = 50.713n
n = 0.00195 mols O2

THEN, use the LRF, limiting reactant formula, to tell you how much product you make.
LRF: what you have, X (molar mass X) (ratio X:Y) (molar mass Y) = Y made or needed for rxn

You get the ratio from the balanced stoichiometry.

4Fe(OH)2 + O2 + 2H2O --> 4Fe(OH)3

Therefore the O2:Fe(OH)3 ratio is 1:4
O2=32g/mol, Fe(OH)3= 107g/mol
http://www.ptable.com

Now it's just PLUG AND CHUG.

0.00195mol O2 (32g/1mol)(4 Fe(OH)3/1 O2)(1mol/107g) = mol Fe(OH)3 made (everything else cancels)

math: 0.00195*32*4/107 = 0.002335 mol Fe(OH)3 made

Easy. Hope this helps! :)

You have a volume a pressure and a temperature. You can use PV = nRT to work out the moles involved and from there work out the end product.