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Since the balanced equation is C3H6S + 6O2 --> 3CO2 + 3H2O + SO3, the ratio is going to be 1:6.

C3H6S: O2:

C: (3x12) + H: (6x1) + S: (1x32) = 74 GFW O: (2x16) = 32 GFW

The number of grams for C3H6S is 61.7g. To find the number of moles you are going to divide 61.7g by 74 GFW. This equals 0.834 moles.

The number of grams for O2 is 240g. To find the number of moles you are going to divide 240g by 32 GFW. This equals 7.5 moles.

Since the ratio that we have already has one of the compounds being one we can leave the ratio as 1:6 C3H6S to O2.

Since you have a 1:6 ratio the number of moles of O2 should be triple that of C3H6S. Since you have 0.834 moles of C3H6S and 7.5 moles of O2, and you are supposed to have a 1:6 ratio, the number of moles for O2 should be 5.004 based on the number of moles of C3H6S (0.834x3). Therefore, you do not have the sufficient amount of C3H6S making it the limiting reagent.

**Question:**For the balanced equation C3H6S + 6O2 --> 3CO2 + 3H2O + SO3, if 61.7 grams of C3H6S is reacted with 240 grams of O2, how many grams of CO2 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 C3H6S + 6O2 --> 3CO2 + 3H2O + SO3, the ratio is going to be 1:6.

**Step Two**: Find the GFW of each compound (do not combine them).C3H6S: O2:

C: (3x12) + H: (6x1) + S: (1x32) = 74 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).__C3H6S:__The number of grams for C3H6S is 61.7g. To find the number of moles you are going to divide 61.7g by 74 GFW. This equals 0.834 moles.

__O2:__The number of grams for O2 is 240g. To find the number of moles you are going to divide 240g by 32 GFW. This equals 7.5 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 1:6 C3H6S 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 ratio the number of moles of O2 should be triple that of C3H6S. Since you have 0.834 moles of C3H6S and 7.5 moles of O2, and you are supposed to have a 1:6 ratio, the number of moles for O2 should be 5.004 based on the number of moles of C3H6S (0.834x3). Therefore, you do not have the sufficient amount of C3H6S 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 is C3H6S and the balanced equation is C3H6S + 6O2 --> 3CO2 + 3H2O + SO3, the ratio between C3H6S and CO2 is a 1:3 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 0.834 moles of C3H6S (^number of moles of C3H6S that we have^) we need a third as many moles of CO2, as shown in the ratio, therefore, we divide the number of moles of C3H6S by 3 to get 0.278 moles of CO2.

**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 CO2: C: (1x12) + O: (2x16)= 44 GFW grams=(0.278moles)(44GFW) ---> grams=12.232g

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