Given the equation 2Fe+3O2=2FeO3 what is the limiting reactant if given 280 grams of Fe and 192 grams os O2?
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the coefficients of a balanced equation give \(molar\) ratios, but the information in the problem has given you a \(mass\) ratio.
What you need to do is find the moles of each reactant using the given mass and its molar mass. Then compare the amounts you \(have\) with the amounts the balanced reaction tells you that you \(need\)
I still do not understand
Find the amount in moles of O2 that you have and the amount of moles of Fe that you have then find how much FeO3 is produced from the amount of O2 and the amount of Fe. Whichever produces less FeO3 is the limiting reactant.
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The stoichiometry of the reaction is telling you that 2 mol of Fe (2Fe) will react with 3 mol of O2 (3O2) completely to produce 2 mol of FeO3 (2FeO3). Thats means that the relationship in moles is 2 mol Fe every 3 mol of O2 to produce 2 mol of FeO3.
Calculate the number of moles of Fe and O2 that you have in 280 grams of Fe and 192 grams of O2.
n= m/ MM
n= number of moles
MM= Molecular mass
MM you have to look in the periodic table for Fe and for O2. Remember the O2 is diatomic then the atomic mass that is in the table has to be multiply by 2 to obtain the molecular mass.
g of Fe x (1 mol Fe/ MM Fe) x (2 mol FeO3 / 2 mol Fe) = A mol of FeO3
g of O2 x (1 mol O2/ MM O2) x (2 mol FeO3 / 3 mol O2) = B mol of FeO3
If A>B then the limiting reactant is B