5. How many grams of iron oxide can be produced from 2.50 g of oxygen reacting with iron, according to the following equation?
2 Fe (s) + 3 O2 (g) -->2 Fe2O3(s)
4 answers
answered
in order to solve this problem, the best way is to calculate their mole ratio(O2 and Fe2O3).so that would be
mole ratio of O2 and Fe2O3;it would be 3:2=2.50g:x..3X=5 so the mole for fe2o3 is 1.67mol....so u got the mole so find the the mass...formula is moles=mass/Mr(Fe2O3)....1.67mol=mass/272g/mol..thus ur answr comes to 454.24g...therefore the mass o iron oxide comes to 454.24grams..
mole ratio of O2 and Fe2O3;it would be 3:2=2.50g:x..3X=5 so the mole for fe2o3 is 1.67mol....so u got the mole so find the the mass...formula is moles=mass/Mr(Fe2O3)....1.67mol=mass/272g/mol..thus ur answr comes to 454.24g...therefore the mass o iron oxide comes to 454.24grams..
No that's not right.
mols O2 = 2.50g/32 = 0.078 but you should do it more accurately.
Convert to mols Fe2O3. That's 0.078 x 2/3 = approx 0.052
Then g Fe2O3 = mols x molar mass = 0.052 x 159.7 = approx 8.3.
mols O2 = 2.50g/32 = 0.078 but you should do it more accurately.
Convert to mols Fe2O3. That's 0.078 x 2/3 = approx 0.052
Then g Fe2O3 = mols x molar mass = 0.052 x 159.7 = approx 8.3.
Calculate the mass of iron needed to completely react with oxygen to produce 5.50 g iron
(TI) oxide.
A Fe + 302
› 2 Fe203
(TI) oxide.
A Fe + 302
› 2 Fe203
2 answers