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Question: For the balanced equation 4FeS2 +11O2 --> 2Fe2O3 + 8SO2, what would be the limiting reagent if 77.3 grams of FeS2 were reacted with 97.3 grams of O2?
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is 4FeS2 +11O2 --> 2Fe2O3 + 8SO2, the ratio is going to be 4:11.
Step Two: Find the GFW of each compound (do not combine them).
FeS2: O2:
Fe: (1x56) + S: (2x32) = 120 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).
FeS2:
The number of grams for FeS2 is 77.3g. To find the number of moles you are going to divide 77.3g by 120 GFW. This equals 0.644 moles.
O2:
The number of grams for O2 is 97.3g. To find the number of moles you are going to divide 97.3g by 32 GFW. This equals 3.041 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 4:11 ratio, we can divide both sides by 4 to make it a 1:2.75 ratio, FeS2 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 1:2.75 ratio the number of moles of O2 should be 2.75 times that of FeS2. Since you have 0.644 moles of FeS2 and 3.041 moles of O2, and you are supposed to have a 1:2.75 ratio, the number of moles for O2 should be 1.771 based on the number of moles of FeS2 (0.644x2.75). Therefore, you have an excess of O2 meaning that you do not have the sufficient amount of FeS2 making it the limiting reagent.
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Question: For the balanced equation 4FeS2 +11O2 --> 2Fe2O3 + 8SO2, what would be the limiting reagent if 77.3 grams of FeS2 were reacted with 97.3 grams of O2?
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is 4FeS2 +11O2 --> 2Fe2O3 + 8SO2, the ratio is going to be 4:11.
Step Two: Find the GFW of each compound (do not combine them).
FeS2: O2:
Fe: (1x56) + S: (2x32) = 120 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).
FeS2:
The number of grams for FeS2 is 77.3g. To find the number of moles you are going to divide 77.3g by 120 GFW. This equals 0.644 moles.
O2:
The number of grams for O2 is 97.3g. To find the number of moles you are going to divide 97.3g by 32 GFW. This equals 3.041 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 4:11 ratio, we can divide both sides by 4 to make it a 1:2.75 ratio, FeS2 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 1:2.75 ratio the number of moles of O2 should be 2.75 times that of FeS2. Since you have 0.644 moles of FeS2 and 3.041 moles of O2, and you are supposed to have a 1:2.75 ratio, the number of moles for O2 should be 1.771 based on the number of moles of FeS2 (0.644x2.75). Therefore, you have an excess of O2 meaning that you do not have the sufficient amount of FeS2 making it the limiting reagent.
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