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Question: For the balanced equation 3Ca(OH)2 + 2H3PO4 --> Ca3(PO4)2 + 6OH, what would be the limiting reagent if 14.0 grams of Ca(OH)2 were reacted with 19.8 grams of H3PO4?
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.
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Question: For the balanced equation 3Ca(OH)2 + 2H3PO4 --> Ca3(PO4)2 + 6OH, what would be the limiting reagent if 14.0 grams of Ca(OH)2 were reacted with 19.8 grams of H3PO4?
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.
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