How many grams of co are needed to react with an excess of fe₂o₃ to produce 156.2 g fe? show your workfe2o3 (s) + 3 co (g) --> 3 co2 (g) + 2 fe (s) - 20

moles Fe = 209.7 g/ 55.847 g/mol=3.75  

the ratio between Fe and CO is 2 : 3  

moles CO = 3.75 x 3 /2 =5.63  

mass CO = 5.63 mol x 28.01 g/mol=157.7 g

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.