On (iii) oxide is formed when iron combines with oxygen in the air. how many moles of fe2o3 are formed when 28.1 g of fe reacts completely with oxygen?

Reaction between Iron and oxygen to form iron (iii) oxide is written as;
4Fe (s) + 3O2(g) = 2Fe2O3(s)
1mole of iron contains 56 g, hence,
28.1 g of iron contains 28.1/56 = 0.502 moles
To get the moles of iron (iii) oxide formed we use the mole ratio
mole ratio of iron : iron (iii) oxide = 4 : 2
Therefore, moles of iron(iii) oxide will be given by 0.502×2/4 = 0.251 moles
hence, the answer is 0.251 moles of iron(iii) oxide

32.06% yield
explanation:
first, calculate theoretical yield of carbonic acid
balanced equation: 2nahco3 → na2co3 + h2co3
step 1: convert 3.24 grams nahco3 to moles nahco3 by dividing by molar mass
          - 3.24 g nahco3 / (22.99 + 1.008 + 12.01 + 3(15.999)) g/mol nahco3 = 0.039         moles
step 2: use mol: mol ratio to convert moles nahco3 to moles h2co3
          - 0.039 mol nahco3 x 1 mol h2co3 / 2 mol nahco3 = 0.0195 moles h2co3
step 3: convert to grams h2co3 by multiplying by molar mass
          - 0.0195 moles h2co3 x ( 2(1.008) + 12.01 + 3(15.999)) = 1.209 grams h2co3
second: calculate the actual yield of carbonic acid, using the (3.24 - 2.19) 1.05 grams nahco3 that actually decomposed.
step 1: convert 1.05 grams nahco3 to moles nahco3 by dividing by molar mass
          - 1.05 g nahco3 / (22.99 + 1.008 + 12.01 + 3(15.999)) g/mol nahco3 = 0.0125 moles
step 2: use mol: mol ratio to convert moles nahco3 to moles h2co3
          - 0.0125 mol nahco3 x 1 mol h2co3 / 2 mol nahco3 = 0.0063 moles h2co3
step 3: convert to grams h2co3 by multiplying by molar mass
          - 0.0063 moles h2co3 x ( 2(1.008) + 12.01 + 3(15.999)) = 0.388 grams h2co3
third: calculate percent yield by dividing actual yield by theoretical yield, and multiplying by 100
0.388 g h2co3 / 1.209 g h2co3 (x 100%) = 32.06% yield.