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Question: For the balanced equation C4H8S2 + 9O2 --> 4CO2 + 4H2O + 2SO3, if 63.1 grams of C4H8S2 is reacted with 209 grams of O2, how many grams of H2O would be produced?
Step One: Find the the limiting reagent.
Here are the steps to finding the limiting reagent
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is C4H8S2 + 9O2 --> 4CO2 + 4H2O + 2SO3, the ratio is going to be 1:9.
Step Two: Find the GFW of each compound (do not combine them).
C4H8S2: O2:
C: (4x12) + H: (8x1) + 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).
C4H8S2:
The number of grams for C4H8S2 is 63.1g. To find the number of moles you are going to divide 63.1g by 120 GFW. This equals 0.526 moles.
O2:
The number of grams for O2 is 209g. To find the number of moles you are going to divide 209g by 32 GFW. This equals 6.531 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 already has one of the compounds being one we can leave the ratio as 1:9, C4H8S2 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 1:9 ratio the number of moles of O2 should be 9 times that of C4H8S2. Since you have 0.526 moles of C4H8S2 and 6.531 moles of O2, and you are supposed to have a 1:9 ratio, the number of moles for O2 should be 4.73 based on the number of moles of C4H8S2 (0.526x9). Therefore, you have an excess of O2 which means that you do not have the sufficient amount of C4H8S2 making it the limiting reagent.
Question: For the balanced equation C4H8S2 + 9O2 --> 4CO2 + 4H2O + 2SO3, if 63.1 grams of C4H8S2 is reacted with 209 grams of O2, how many grams of H2O would be produced?
Step One: Find the the limiting reagent.
Here are the steps to finding the limiting reagent
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is C4H8S2 + 9O2 --> 4CO2 + 4H2O + 2SO3, the ratio is going to be 1:9.
Step Two: Find the GFW of each compound (do not combine them).
C4H8S2: O2:
C: (4x12) + H: (8x1) + 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).
C4H8S2:
The number of grams for C4H8S2 is 63.1g. To find the number of moles you are going to divide 63.1g by 120 GFW. This equals 0.526 moles.
O2:
The number of grams for O2 is 209g. To find the number of moles you are going to divide 209g by 32 GFW. This equals 6.531 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 already has one of the compounds being one we can leave the ratio as 1:9, C4H8S2 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 1:9 ratio the number of moles of O2 should be 9 times that of C4H8S2. Since you have 0.526 moles of C4H8S2 and 6.531 moles of O2, and you are supposed to have a 1:9 ratio, the number of moles for O2 should be 4.73 based on the number of moles of C4H8S2 (0.526x9). Therefore, you have an excess of O2 which means that you do not have the sufficient amount of C4H8S2 making it the limiting reagent.
Step Two: Use the limiting reagent and see what the ratio is between that and the compound in the product (this is what you are trying to figure out). Make this ratio 1:? or ?:1
Since the limiting reagent isC4H8S2and the balanced equation is C4H8S2 + 9O2 --> 4CO2 + 4H2O + 2SO3, the ratio between C4H8S2 and H2O is a 1:4 ratio.
Step Three: Using the ratio between the limiting reagent and the compound in the product, find the number of moles in the product's compound by multiplying/diving the limiting reagent by the non-one number in the ratio (1:?).
For every 0.526 moles of C4H8S2 (^number of moles of C4H8S2 that we have^) we need four times as many moles of H2O, as shown in the ratio, therefore, we multiply the number of moles of C4H8S2 by 4 to get 2.104 moles of H2O.
Step Four: Once you know how many moles of the compound in the product you have, turn the moles back into grams. This allows you to answer the question of how much of ___ compound is able to be produced.
To turn the moles back into grams we used the equation grams=(moles)(GFW).
GFW of H2O: H: (2x1) + O: (1x16) = 18 GFW grams=(2.104moles)(18GFW) ---> grams=37.872g
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Since the limiting reagent isC4H8S2and the balanced equation is C4H8S2 + 9O2 --> 4CO2 + 4H2O + 2SO3, the ratio between C4H8S2 and H2O is a 1:4 ratio.
Step Three: Using the ratio between the limiting reagent and the compound in the product, find the number of moles in the product's compound by multiplying/diving the limiting reagent by the non-one number in the ratio (1:?).
For every 0.526 moles of C4H8S2 (^number of moles of C4H8S2 that we have^) we need four times as many moles of H2O, as shown in the ratio, therefore, we multiply the number of moles of C4H8S2 by 4 to get 2.104 moles of H2O.
Step Four: Once you know how many moles of the compound in the product you have, turn the moles back into grams. This allows you to answer the question of how much of ___ compound is able to be produced.
To turn the moles back into grams we used the equation grams=(moles)(GFW).
GFW of H2O: H: (2x1) + O: (1x16) = 18 GFW grams=(2.104moles)(18GFW) ---> grams=37.872g
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