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Since the balanced equation is 2C6H6S2 + 13O2 --> 12CO + 6H2O + 4SO2, the ratio is going to be 2:13.

C6H6S2: O2:

C: (6x12) + H: (6x1) + S: (2x32)= 142 GFW O: (2x16) = 32 GFW

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.

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.

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.

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.

**Question:**For the balanced equation 2C6H6S2 + 13O2 --> 12CO + 6H2O + 4SO2, if 8.36 grams of C6H6S2 is reacted with 21.3 grams of O2, how many grams of CO 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 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.

**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 C6H6S2 and the balanced equation is 2C6H6S2 + 13O2 --> 12CO + 6H2O + 4SO2, the ratio between C6H6S2 and CO is a 2:12 ratio or a 1:6 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 0.059 moles of C6H6S2 (^number of moles of C6H6S2 that we have^) we need six times as many moles of CO, as shown in the ratio, therefore, we multiply the number of moles of C6H6S2 by 6 to get 0.354 moles of CO.

**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 CO: C: (1x12) + O: (1x16) = 28 GFW grams=(0.354moles)(28GFW) ---> grams=9.912g

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