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
Well, now that you've calculated the molar masses of each compound, you can use them to calculate the moles of each substance.
FeS: 0.036
H2S: 0.047
CO2: 0.363
Next, you need to determine the limiting reagent. To do this, compare the moles of each substance in the given amounts to the stoichiometric coefficients of the balanced equation. Let's assume the balanced equation is:
FeS + H2S + CO2 → HCO2H + Other products
Since we have a 47.0% yield, let's calculate the moles of HCO2H that would be obtained if the reaction went to completion.
HCO2H: 0.363 x (1 mole HCO2H / 1 mole CO2) = 0.363 moles of HCO2H
Now, compare the moles of HCO2H obtained if the reaction went to completion with the moles of HCO2H obtained from each reactant.
FeS: 0.036 x (1 mole HCO2H / 1 mole FeS) = 0.036 moles of HCO2H
H2S: 0.047 x (1 mole HCO2H / 1 mole H2S) = 0.047 moles of HCO2H
The smallest value obtained is 0.036 moles from FeS. Therefore, FeS is the limiting reagent. This means that all the FeS will be consumed in the reaction, and the amount of HCO2H obtained will depend on the stoichiometry.
So, if the reaction went to completion, you would obtain 0.036 moles of HCO2H. However, the reaction only yields 47.0% of the theoretical maximum, so you need to multiply the value by 0.47:
0.036 moles of HCO2H x 0.47 = 0.01692 moles of HCO2H
Finally, multiply the moles of HCO2H by its molar mass to find the mass of HCO2H obtained:
0.01692 moles of HCO2H x 46.03 g/mol (molar mass of HCO2H) = 0.779 g of HCO2H (rounded to 3 significant figures)
Therefore, approximately 0.779 grams of HCO2H would be obtained.
After calculating the molar masses of FeS, H2S, and CO2, you need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed and determines the amount of product that can be formed.
To find the limiting reagent, compare the molar ratios of the reactants to the coefficients in the balanced equation. The balanced equation for the reaction is:
FeS + 2 H2S + 2 CO2 → Fe(HCO2)2 + 2 S
According to the balanced equation, the molar ratio between FeS and Fe(HCO2)2 is 1:1. The molar ratio between H2S and Fe(HCO2)2 is 2:1, and the molar ratio between CO2 and Fe(HCO2)2 is 2:1.
To compare the molar ratios to the given amounts, divide the moles of each reactant by their respective stoichiometric coefficients:
FeS: 0.036 mol / 1 = 0.036 mol
H2S: 0.047 mol / 2 = 0.0235 mol
CO2: 0.363 mol / 2 = 0.1815 mol
Since the molar ratio between FeS and Fe(HCO2)2 is 1:1, FeS is the limiting reagent since it has the smallest number of moles. This means that all of the FeS will be consumed, and the amount of Fe(HCO2)2 formed will be determined by the number of moles of FeS.
To calculate the amount of Fe(HCO2)2 formed, you need to convert the moles of FeS to moles of Fe(HCO2)2. Since the ratio between FeS and Fe(HCO2)2 is 1:1, you can simply use the number of moles of FeS:
Fe(HCO2)2: 0.036 mol
Finally, to obtain the actual amount of Fe(HCO2)2 obtained, multiply the number of moles by the molar mass of Fe(HCO2)2:
Fe(HCO2)2: 0.036 mol * (146.1 g/mol) = 5.26 g
Therefore, 5.26g of Fe(HCO2)2 is obtained in the reaction. However, since the reaction is stated to have a 47.0% yield, the actual amount of Fe(HCO2)2 obtained will be:
Actual yield = 47.0% * 5.26 g = 2.4742 g (rounded to 4 significant figures)
So, the amount of Fe(HCO2)2 obtained would be approximately 2.47 g.
After finding the molar mass of each compound, the next step is to determine the balanced chemical equation for the reaction and calculate the limiting reactant.
The balanced chemical equation for the reaction between FeS, H2S, and CO2 can be written as:
FeS + 2H2S + CO2 → Fe(HCO2)2 + 2S
From this equation, we can see that the stoichiometric ratio between FeS and Fe(HCO2)2 is 1:1. This means that 1 mole of FeS reacts to produce 1 mole of Fe(HCO2)2.
To calculate the limiting reactant, we need to compare the number of moles of each reactant with their respective stoichiometric ratios.
For FeS:
Number of moles of FeS = Mass of FeS / Molar mass of FeS
= 3.0 g / 87.9 g/mol
≈ 0.034 moles
For H2S:
Number of moles of H2S = Mass of H2S / Molar mass of H2S
= 1.60 g / 34.1 g/mol
≈ 0.047 moles
For CO2:
Number of moles of CO2 = Mass of CO2 / Molar mass of CO2
= 1.60 g / 44.0 g/mol
≈ 0.036 moles
Since the stoichiometric ratio is 1:1 for FeS and Fe(HCO2)2, we can see that FeS is the limiting reactant as it has the lowest number of moles.
To find the moles of Fe(HCO2)2 produced, we multiply the number of moles of FeS by the yield percentage:
Moles of Fe(HCO2)2 = Moles of FeS × Yield percentage
= 0.034 moles × 0.47
≈ 0.016 moles
Finally, to find the mass of Fe(HCO2)2 produced, we use the molar mass of Fe(HCO2)2 and the number of moles:
Mass of Fe(HCO2)2 = Moles of Fe(HCO2)2 × Molar mass of Fe(HCO2)2
= 0.016 moles × (56 + 1 + 16 + 16 + 1 + 1) g/mol
≈ 1.4 g
Therefore, approximately 1.4 g of Fe(HCO2)2 is obtained in the reaction.
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.