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Question: For the balanced equation 2C6H6S2 + 13O2 --> 12CO + 6H2O + 4SO2, what would be the limiting reagent if 8.36 grams of C6H6S2 were reacted with 21.3 grams of O2?
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is 2C6H6S2 + 13O2 --> 12CO + 6H2O + 4SO2, the ratio is going to be 2:13.
Step Two: Find the GFW of each compound (do not combine them).
C6H6S2: O2:
C: (6x12) + H: (6x1) + S: (2x32)= 142 GFW O: (2x16) = 32 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).
C6H6S2:
The number of grams for C6H6S2 is 8.36g. To find the number of moles you are going to divide 8.36g by 142 GFW. This equals 0.059 moles.
O2:
The number of grams for O2 is 21.3g. To find the number of moles you are going to divide 21.3g by 32 GFW. This equals 0.666 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.
Since the ratio that we have is a 2:13 ratio, we can divide both sides by 2 to make it a 1:6.5 ratio, C6H6S2 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 1:6.5 ratio the number of moles of O2 should be 6.5 times that of C6H6S2. Since you have 0.059 moles of C6H6S2 and 0.666 moles of O2, and you are supposed to have a 1:6.5 ratio, the number of moles for O2 should be 0.384 based on the number of moles of C6H6S2 (0.0.59x6.5). Therefore, you have an excess of O2 meaning that you do not have the sufficient amount of C6H6S2 making it the limiting reagent.
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Question: For the balanced equation 2C6H6S2 + 13O2 --> 12CO + 6H2O + 4SO2, what would be the limiting reagent if 8.36 grams of C6H6S2 were reacted with 21.3 grams of O2?
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is 2C6H6S2 + 13O2 --> 12CO + 6H2O + 4SO2, the ratio is going to be 2:13.
Step Two: Find the GFW of each compound (do not combine them).
C6H6S2: O2:
C: (6x12) + H: (6x1) + S: (2x32)= 142 GFW O: (2x16) = 32 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).
C6H6S2:
The number of grams for C6H6S2 is 8.36g. To find the number of moles you are going to divide 8.36g by 142 GFW. This equals 0.059 moles.
O2:
The number of grams for O2 is 21.3g. To find the number of moles you are going to divide 21.3g by 32 GFW. This equals 0.666 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.
Since the ratio that we have is a 2:13 ratio, we can divide both sides by 2 to make it a 1:6.5 ratio, C6H6S2 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 1:6.5 ratio the number of moles of O2 should be 6.5 times that of C6H6S2. Since you have 0.059 moles of C6H6S2 and 0.666 moles of O2, and you are supposed to have a 1:6.5 ratio, the number of moles for O2 should be 0.384 based on the number of moles of C6H6S2 (0.0.59x6.5). Therefore, you have an excess of O2 meaning that you do not have the sufficient amount of C6H6S2 making it the limiting reagent.
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