How many grams of CO are needed with an excess of Fe2O3 to produce 35.0 g Fe? I just don't get how to set up so I could do the work
Here is an example stoichiometry problem. Look at it, follow the steps that apply to your question. I'll be glad to answer questions you don't understand.
Sorry, here is the link.
https://www.jiskha.com/science/chemistry/stoichiometry.html
https://www.jiskha.com/display.cgi?id=1515471953
To solve this problem, we will use the concept of stoichiometry, which involves using balanced chemical equations to relate the amounts of reactants and products. Here's how you can set up the problem:
Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction between CO and Fe2O3 is:
Fe2O3 + 3CO -> 2Fe + 3CO2
This equation tells us that three moles of CO are required to react with one mole of Fe2O3 to produce two moles of Fe and three moles of CO2.
Step 2: Determine the molar mass of Fe.
The molar mass of Fe is 55.85 g/mol. This means that one mole of Fe weighs 55.85 grams.
Step 3: Convert the given mass of Fe to moles.
To determine the amount of Fe produced, we need to convert the given mass of 35.0 g Fe to moles. To do this, divide the mass of Fe by its molar mass:
35.0 g Fe / 55.85 g/mol = 0.626 moles Fe
Step 4: Use stoichiometry to relate the moles of Fe to moles of CO.
According to the balanced equation, three moles of CO are required to produce two moles of Fe. This means that:
3 moles CO = 2 moles Fe
Using this relationship, we can set up a proportion to find the moles of CO:
(3 moles CO / 2 moles Fe) = (x moles CO / 0.626 moles Fe)
Solving for x, we get:
x = (3 moles CO / 2 moles Fe) * 0.626 moles Fe
x = 0.939 moles CO
Step 5: Convert moles of CO to grams.
Finally, we need to convert the moles of CO to grams. To do this, we multiply the number of moles by the molar mass of CO, which is 28.01 g/mol:
0.939 moles CO * 28.01 g/mol = 26.33 g CO
Therefore, approximately 26.33 grams of CO are needed to produce 35.0 grams of Fe with an excess of Fe2O3.