How to find the THeoretical yield
Step One: Find 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
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/dividing the limiting reagent by the non-one number in the ratio (1:?)
Step Four: Once you know how many moles of the compound in the product you have, turn the moles back into gram. This allows you to answer the question of how much of ___ compound is able to be produced
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
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/dividing the limiting reagent by the non-one number in the ratio (1:?)
Step Four: Once you know how many moles of the compound in the product you have, turn the moles back into gram. This allows you to answer the question of how much of ___ compound is able to be produced
3 worked out examples of how to find the theoretical yield
1) For the balanced equation below, if 1 gram of H2 is reacted with 8 grams of O2, how many grams of H2O would be produced.
2H2 + O2 --> 2H2O
Step One: Find the the limiting reagent.
There is no limiting reagent because there is 0.5 moles of H2 (1g/2GFW) and 0.25 moles of O2 (8g/16GFW) with a 2:1 ratio.
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:?
Since there is no limiting reagent you can choose either H2 or O2. For this scenario we are going to use H2. H2 has a 2:2 or 1:1 ratio with H2O.
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:?).
Since there is a 1:1 ratio between H2 and H2O, the number of moles in H2O is the same as the number of moles in H2 (0.5 moles).
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.
The GFW of H2O is 18 (2H-2 + O-16) and the number of moles is 0.5. You then plug this into the equation (grams)=(moles)(GFW) and get that 9 grams of H2O are able to be produced.
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2) For the balanced equation shown below, if 87.6 grams of Al were reacted with 130 grams of Cr2O3, how many grams of Cr would be produced?
2Al + Cr2O3 --> Al2O3 + 2Cr
Step One: Find the the limiting reagent.
Cr2O3 is the limiting reagent because you only have 0.86 moles of it (130g/152GFW) and you have 3.24 moles of Al (87.6g/27GFW) with a 1:2 ratio between Cr2O3 and Al. For there not to be a limiting reagent you would either need for there to be 1.62 moles of Cr2O3 or 1.72 moles of Al.
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:?
Cr2O3 has a 1:2 ratio with Cr.
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:?).
Since there is a 1:2 ratio between Cr2O3 and Cr, you have to double the number of moles in Cr2O3 to find the number of moles in Cr. 0.86 moles x 2 = 1.72 moles of Cr.
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.
The GFW of Cr is 52 (1Cr-52) and the number of moles is 1.72. You then plug this into the equation (grams)=(moles)(GFW) and get that 89.44 grams of Cr are able to be produced.
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3) For the balanced equation shown below, if 9.12 grams of NH3 were reacted with 13.1 grams of O2, how many grams of H2O would be produced?
4NH3 + 5O2 --> 6H2O + 4NO
Step One: Find the the limiting reagent.
O2 is the limiting reagent because you only have 0.41 moles of it (13.1g/32GFW) and you have 0.54 moles of NH3 (9.12g/17GFW) with a 5:4 ratio between O2 and NH3. For there not to be a limiting reagent you would either need for there to be 0.675 moles of O2 or 0.328 moles of NH3.
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:?
O2 has a 5:6 ratio with H2O or a 1:1.2 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:?).
Since there is a 1:1.2 ratio between O2 and H2O, you have to multiply the number of moles in O2 by 1.2 to find the number of moles in H2O. 0.41 moles x 1.2 = 0.492 moles of H2O.
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.
The GFW of H2O is 18 (2H-2 + O-16) and the number of moles is 0.492. You then plug this into the equation (grams)=(moles)(GFW) and get that 8.856 grams of H2O are able to be produced.
2H2 + O2 --> 2H2O
Step One: Find the the limiting reagent.
There is no limiting reagent because there is 0.5 moles of H2 (1g/2GFW) and 0.25 moles of O2 (8g/16GFW) with a 2:1 ratio.
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:?
Since there is no limiting reagent you can choose either H2 or O2. For this scenario we are going to use H2. H2 has a 2:2 or 1:1 ratio with H2O.
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:?).
Since there is a 1:1 ratio between H2 and H2O, the number of moles in H2O is the same as the number of moles in H2 (0.5 moles).
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.
The GFW of H2O is 18 (2H-2 + O-16) and the number of moles is 0.5. You then plug this into the equation (grams)=(moles)(GFW) and get that 9 grams of H2O are able to be produced.
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2) For the balanced equation shown below, if 87.6 grams of Al were reacted with 130 grams of Cr2O3, how many grams of Cr would be produced?
2Al + Cr2O3 --> Al2O3 + 2Cr
Step One: Find the the limiting reagent.
Cr2O3 is the limiting reagent because you only have 0.86 moles of it (130g/152GFW) and you have 3.24 moles of Al (87.6g/27GFW) with a 1:2 ratio between Cr2O3 and Al. For there not to be a limiting reagent you would either need for there to be 1.62 moles of Cr2O3 or 1.72 moles of Al.
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:?
Cr2O3 has a 1:2 ratio with Cr.
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:?).
Since there is a 1:2 ratio between Cr2O3 and Cr, you have to double the number of moles in Cr2O3 to find the number of moles in Cr. 0.86 moles x 2 = 1.72 moles of Cr.
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.
The GFW of Cr is 52 (1Cr-52) and the number of moles is 1.72. You then plug this into the equation (grams)=(moles)(GFW) and get that 89.44 grams of Cr are able to be produced.
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3) For the balanced equation shown below, if 9.12 grams of NH3 were reacted with 13.1 grams of O2, how many grams of H2O would be produced?
4NH3 + 5O2 --> 6H2O + 4NO
Step One: Find the the limiting reagent.
O2 is the limiting reagent because you only have 0.41 moles of it (13.1g/32GFW) and you have 0.54 moles of NH3 (9.12g/17GFW) with a 5:4 ratio between O2 and NH3. For there not to be a limiting reagent you would either need for there to be 0.675 moles of O2 or 0.328 moles of NH3.
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:?
O2 has a 5:6 ratio with H2O or a 1:1.2 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:?).
Since there is a 1:1.2 ratio between O2 and H2O, you have to multiply the number of moles in O2 by 1.2 to find the number of moles in H2O. 0.41 moles x 1.2 = 0.492 moles of H2O.
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.
The GFW of H2O is 18 (2H-2 + O-16) and the number of moles is 0.492. You then plug this into the equation (grams)=(moles)(GFW) and get that 8.856 grams of H2O are able to be produced.