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Question: For the balanced equation 3HCl + Na3PO4 --> H3PO4 + 3NaCl, if 24.0 grams of HCl is reacted with 59.8 grams of Na3PO4, how many grams of NaCl 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 3HCl + Na3PO4 --> H3PO4 + 3NaCl, the ratio is going to be 3:1.
Step Two: Find the GFW of each compound (do not combine them).
HCl: Na3PO4:
H: (1x1) + Cl: (1x35.5) = 36.5 GFW Na: (3x23) + P: (1x31) + O: (4x16) = 164 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).
HCl:
The number of grams for HCl is 24.0g. To find the number of moles you are going to divide 24.0g by 36.5 GFW. This equals 0.658 moles.
Na3PO4:
The number of grams for Na3PO4 is 59.8g. To find the number of moles you are going to divide 59.8g by 164 GFW. This equals 0.365 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 3:1 HCl to Na3PO4.
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 3:1 ratio the number of moles of HCl should be triple that of Na3PO4. Since you have 0.658 moles of HCl and 0.365 moles of Na3PO4, and you are supposed to have a 3:1 ratio, the number of moles for HCl should be 10.95 based on the number of moles of Na3PO4 (0.365x3). Therefore, you do not have the sufficient amount of HCl making it the limiting reagent.
Question: For the balanced equation 3HCl + Na3PO4 --> H3PO4 + 3NaCl, if 24.0 grams of HCl is reacted with 59.8 grams of Na3PO4, how many grams of NaCl 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 3HCl + Na3PO4 --> H3PO4 + 3NaCl, the ratio is going to be 3:1.
Step Two: Find the GFW of each compound (do not combine them).
HCl: Na3PO4:
H: (1x1) + Cl: (1x35.5) = 36.5 GFW Na: (3x23) + P: (1x31) + O: (4x16) = 164 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).
HCl:
The number of grams for HCl is 24.0g. To find the number of moles you are going to divide 24.0g by 36.5 GFW. This equals 0.658 moles.
Na3PO4:
The number of grams for Na3PO4 is 59.8g. To find the number of moles you are going to divide 59.8g by 164 GFW. This equals 0.365 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 3:1 HCl to Na3PO4.
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 3:1 ratio the number of moles of HCl should be triple that of Na3PO4. Since you have 0.658 moles of HCl and 0.365 moles of Na3PO4, and you are supposed to have a 3:1 ratio, the number of moles for HCl should be 10.95 based on the number of moles of Na3PO4 (0.365x3). Therefore, you do not have the sufficient amount of HCl 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 HCl and the balanced equation is 3HCl + Na3PO4 --> H3PO4 + 3NaCl, the ratio between HCl and NaCl is a 3:3 or 1:1 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 0.658 moles of HCl (^number of moles of HCl that we have^) we need the same number of moles of NaCl, as shown in the ratio, therefore, the number of moles of NaCl is 0.658.
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 NaCl: Na: (1x23) + Cl: (1x35.5) = 58.5 GFW grams=(0.658moles)(58.5GFW) ---> grams=38.49g
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Since the limiting reagent is HCl and the balanced equation is 3HCl + Na3PO4 --> H3PO4 + 3NaCl, the ratio between HCl and NaCl is a 3:3 or 1:1 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 0.658 moles of HCl (^number of moles of HCl that we have^) we need the same number of moles of NaCl, as shown in the ratio, therefore, the number of moles of NaCl is 0.658.
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 NaCl: Na: (1x23) + Cl: (1x35.5) = 58.5 GFW grams=(0.658moles)(58.5GFW) ---> grams=38.49g
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