Step 1. Write and balance the equation.
This is a limiting reagent problem but more complicated than most because it has three reactants instead of the usual two. I find the easiest way to do these (others have other ways of doing it) is to take each reactant individually and ask myself how much product I could get using the amount in the problem and all of the other reagents I need. The first step you have done which is to convert grams to moles.
Step2a. Using the coefficients in the balanced equation, convert moles FeS to moles HCO2H.
2b. Same procedure for H2S.
2c. Same procedure for CO2.
Step 3. The way this usually works is you obtain three different answers for moles HCO2H from 2a, 2b, and 2c. Of course only one answer can be correct; the correct answer in limiting reagent problems is ALWAYS the smallest value and the reactant providing that number is the limiting reagent.
Step 4. Using the smallest number of moles formed, convert to grams. g = mols x molar mass. This is the theoretical yield in grams if yield is 100%.
Step 5. Since this is not 100% (but 47%), multiply theoretical yield x 0.47 to convert to actual yield.
How much HCO2H is obtained by reacting 3.0g of FeS, 1.60g H2S, and 1.60g CO2 if the reaction results in a 47.0% yield?
I know that I need to the molar mass of each which
FeS-3.2/87.9=0.036
H2S-1.6/34.1=0.047
CO2-1.6/44.0=0.363
I just don't know other step I have to go after this one
1 answer
1 answer