i dont understand stoichiometry, please help

hey

The equation for one of the reactions in the process of reducing iron ore to the metal is 

Fe2O3(s) + 3 co (g) = 2Fe (s) + 3CO2 (g) 

what is the maximum mass of iron, in grams, that can be obtained from 454g of iron (iii) oxide?

how do i start this stoichiometry problem?

For this type of problem it is important to first take a step back and understand what they are asking. Given an amount of one of the reactants, they are asking for an amount of one of the products. In order to go from reactants to products all you have to do is use the coefficients in front of the compounds to convert from mol of the reactant to mol of the product. Be sure to have a balanced equation before you start. Also, It is important to realize that you can not compare apples to oranges, meaning you have to first convert the grams to moles, then use the moles to compare to moles of the product, and then simply convert moles of the product to grams of the product because that is what the question is asking for. Simply do dimensional analysis starting with what they give you, and make sure you end with what they are looking for. Attached is the worked out solution. 

Keep in mind that if they gave you grams for each of the reactants, this would now be a limiting reactant question and you would first have to determine the limiting factor before doing this dimensional analysis, but in this case they only gave us a mass of one reactant. 

 

Hope that helps,

Michael Rosen

Author of "The Guide to Surviving General Chemistry"

www.survivechem.com

Insert Image: 

Hope that Helps,

Michael Rosen - Author of "The Guide to Surviving General Chemistry"

www.survivechem.com

The equation is already balanced for you. Next all you have to do is use the mole road ,if you teacher gave you one or explained what it was, to convert the moles of unknown , then you use mole ratio to convert moles of known to moles unknown finally you use the mole road again to convert moles of unknown to your desired units.