How many grams of CO are needed to react with an excess of Fe2O3 to produce 233.4 gFe?
Fe2O3 (s) + 3CO (g) --> 3CO2 (g) + 2Fe (s)
Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
344g
To determine the amount of CO needed, we can use the concept of stoichiometry, which relates the balanced chemical equation to the moles of the reactants and products.
1. Start by finding the moles of Fe using the given mass of Fe (233.4 g) and the molar mass of Fe (55.845 g/mol):
Moles of Fe = Mass of Fe / Molar mass of Fe
= 233.4 g / 55.845 g/mol
2. The balanced chemical equation tells us that the ratio between Fe2O3 and Fe is 1:2. Therefore, the moles of Fe2O3 can be determined by multiplying the moles of Fe by a ratio of 1:2:
Moles of Fe2O3 = Moles of Fe × (1 mol Fe2O3 / 2 mol Fe)
3. The balanced chemical equation also tells us that the ratio between Fe2O3 and CO is 1:3. Therefore, the moles of CO can be determined by multiplying the moles of Fe2O3 by a ratio of 1:3:
Moles of CO = Moles of Fe2O3 × (3 mol CO / 1 mol Fe2O3)
4. Finally, calculate the mass of CO needed using the moles of CO and the molar mass of CO (28.01 g/mol):
Mass of CO = Moles of CO × Molar mass of CO
Plug in the values into the equations to calculate the mass of CO needed.