Question #a4ff9

Start by writing the balanced chemical equation for your reaction

#2Fe_((s)) + O_(2(g)) -> color(red)(2)FeO_((s))#

Notice that you have a #1:color(red)(2)# mole ratio between oxygen gas and iron (II) oxide. This means that, regardless of how many moles of oxygen gas react, the reaction will always produce twice as many moles of iron (II) oxygen.

Since you know that iron is in excess, you can safely assume that all the available oxygen will indeed react to form the oxide.

Use oxygen gas' molar mass to determine how many moles reacted

#29.3cancel("g") * ("1 mole "O_2)/(32.0cancel("g")) = "0.9156 moles"# #O_2#

This means that the reaction produced

#0.9156cancel("moles"O_2) * (color(red)(2)" moles "FeO)/(1cancel("mole"O_2)) = "1.831 moles"# #FeO#

Now all you have to do is use iron (II) oxide's molar mass to determine how many grams would contain this many moles.

#1.831cancel("moles"FeO) * "71.84 g"/(1cancel("mole"FeO)) = "131.54 g"#

Rounded to three sig figs, the number of sig figs you gave for the mass of oxygen gas, the answer will be

#m_(FeO) = color(green)("132 g")#