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Question: For the balanced equation C6H6OS + 8O2 --> 6CO2 + 3H2O + SO2, what would be the limiting reagent if 96.2 grams of C6H6OS were reacted with 120 grams of O2?
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is C6H6OS + 8O2 --> 6CO2 + 3H2O + SO2, the ratio is going to be 1:8.
Step Two: Find the GFW of each compound (do not combine them).
C6H6OS: O2:
C: (6x12) + H: (6x1) + O: (1x16) + S: (1x32) = 126 GFW O: (2x16) = 32 GFWO: (1x16) + S: (1x32)= 126 GFW
Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).
C6H6OS:
The number of grams for C6H6OS is 96.2g. To find the number of moles you are going to divide 96.2g by 126 GFW. This equals 0.763 moles.
O2:
The number of grams for O2 is 120g. To find the number of moles you are going to divide 120g by 32 GFW. This equals 3.75 moles.
Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.
SSince the ratio that we have already has one of the compounds being one we can leave the ratio as 1:8 C6H6OS to O2..
Step Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.
Since you have a 1:8 ratio the number of moles of O2 should be 8 times that of C6H6OS. Since you have 0.763 moles of C6H6OS and 3.75 moles of O2, and you are supposed to have a 1:8 ratio, the number of moles for O2 should be 6.108 based on the number of moles of C6H6OS (0.763x8). Therefore, you do not have the sufficient amount of O2 making it the limiting reagent.
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Question: For the balanced equation C6H6OS + 8O2 --> 6CO2 + 3H2O + SO2, what would be the limiting reagent if 96.2 grams of C6H6OS were reacted with 120 grams of O2?
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is C6H6OS + 8O2 --> 6CO2 + 3H2O + SO2, the ratio is going to be 1:8.
Step Two: Find the GFW of each compound (do not combine them).
C6H6OS: O2:
C: (6x12) + H: (6x1) + O: (1x16) + S: (1x32) = 126 GFW O: (2x16) = 32 GFWO: (1x16) + S: (1x32)= 126 GFW
Step Three: Find the number of moles for each compound. You can do this by dividing the number of grams for one of the compounds by its GFW (do this for both compounds, separately).
C6H6OS:
The number of grams for C6H6OS is 96.2g. To find the number of moles you are going to divide 96.2g by 126 GFW. This equals 0.763 moles.
O2:
The number of grams for O2 is 120g. To find the number of moles you are going to divide 120g by 32 GFW. This equals 3.75 moles.
Step Four: Take the ratio and rewrite it so that it’s a one to something ratio (1:?). This is done by dividing both numbers in the ratio by the smaller number in the ratio.
SSince the ratio that we have already has one of the compounds being one we can leave the ratio as 1:8 C6H6OS to O2..
Step Five: Whichever compound is the “one part” of the ratio, multiply its number of moles by the other number in the ratio. If the product is greater than the number of moles in the second compound, the second compound is the limiting reagent. If the product is less than the number of moles in the second compound, the first compound is the limiting reagent.
Since you have a 1:8 ratio the number of moles of O2 should be 8 times that of C6H6OS. Since you have 0.763 moles of C6H6OS and 3.75 moles of O2, and you are supposed to have a 1:8 ratio, the number of moles for O2 should be 6.108 based on the number of moles of C6H6OS (0.763x8). Therefore, you do not have the sufficient amount of O2 making it the limiting reagent.
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