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Since the balanced equation is 3Ca(OH)2 + 2H3PO4 --> Ca3(PO4)2 + 6OH, the ratio is going to be 3:2.

Ca(OH)2: H3PO4:

Ca: (1x40) + O: (2x16) + H: (2x1)= 74 GFW H: (3x1) + P: (1x31) + O: (4x16) = 98 GFW

The number of grams for Ca(OH)2 is 14.0g. To find the number of moles you are going to divide 14.0g by 74 GFW. This equals 0.189 moles.

The number of grams for H3PO4 is 19.8g. To find the number of moles you are going to divide 19.8g by 98 GFW. This equals 0.202 moles.

Since the ratio that we have is a 3:2 ratio, we can divide both sides by 2 to make it a 1.5:1 ratio, Ca(OH)2 to H3PO4.

Since you have a 1.5:1 ratio the number of moles of Ca(OH)2 should be 1.5 times that of H3PO4. Since you have 0.189 moles of Ca(OH)2 and 0.202 moles of H3PO4, and you are supposed to have a 1.5:1 ratio, the number of moles for Ca(OH)2 should be 0.303 based on the number of moles of H3PO4 (0.202x1.5). Therefore, you do not have the sufficient amount of Ca(OH)2 making it the limiting reagent.

**Question:**For the balanced equation 3Ca(OH)2 + 2H3PO4 --> Ca3(PO4)2 + 6OH, if 14.0 grams of Ca(OH)2 is reacted with 19.8 grams of H3PO4, how many grams of OH 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 3Ca(OH)2 + 2H3PO4 --> Ca3(PO4)2 + 6OH, the ratio is going to be 3:2.

**Step Two**: Find the GFW of each compound (do not combine them).Ca(OH)2: H3PO4:

Ca: (1x40) + O: (2x16) + H: (2x1)= 74 GFW H: (3x1) + P: (1x31) + O: (4x16) = 98 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).__Ca(OH)2:__The number of grams for Ca(OH)2 is 14.0g. To find the number of moles you are going to divide 14.0g by 74 GFW. This equals 0.189 moles.

__H3PO4:__The number of grams for H3PO4 is 19.8g. To find the number of moles you are going to divide 19.8g by 98 GFW. This equals 0.202 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 3:2 ratio, we can divide both sides by 2 to make it a 1.5:1 ratio, Ca(OH)2 to H3PO4.

**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 1.5:1 ratio the number of moles of Ca(OH)2 should be 1.5 times that of H3PO4. Since you have 0.189 moles of Ca(OH)2 and 0.202 moles of H3PO4, and you are supposed to have a 1.5:1 ratio, the number of moles for Ca(OH)2 should be 0.303 based on the number of moles of H3PO4 (0.202x1.5). Therefore, you do not have the sufficient amount of Ca(OH)2 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 Ca(OH)2 and the balanced equation is 3Ca(OH)2 + 2H3PO4 --> Ca3(PO4)2 + 6OH, the ratio between Ca(OH)2 and OH is a 3:6 ratio or a 1:2 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 0.189 moles of Ca(OH)2 (^number of moles of Ca(OH)2 that we have^) we need twice as many moles of OH, as shown in the ratio, therefore, we multiply the number of moles of Ca(OH)2 by 2 to get 0.378 moles of OH.

**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 OH: O: (1x16) + H: (1x1)= 17 GFW grams=(0.378moles)(17GFW) ---> grams=6.426g

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