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Question: For the balanced equation 2HNO3 + NO --> 3NO2 + H2O, if 98.1 grams of HNO3 is reacted with 18.4 grams of NO, how many grams of NO2 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 2HNO3 + NO --> 3NO2 +H2O, the ratio is going to be 2:1.
Step Two: Find the GFW of each compound (do not combine them).
HNO3: NO:
H: (1x1) + N: (1x14) + O: (3x16) = 63 GFW N: (1x14) + O: (1x16) = 30 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).
HNO3:
The number of grams for HNO3 is 98.1g. To find the number of moles you are going to divide 98.1g by 63 GFW. This equals 1.557 moles.
NO:
The number of grams for NO is 18.4g. To find the number of moles you are going to divide 18.4g by 30 GFW. This equals 0.613 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 2:1 HNO3 to NO.
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 2:1 ratio the number of moles of HNO3 should be double that of NO. Since you have 1.557 moles of HNO3 and 0.613 moles of NO, and you are supposed to have a 2:1 ratio, the number of moles for HNO3 should be 1.226 based on the number of moles of NO (0.613x2). Therefore, you have an excess of HNO3 meaning that you do not have the sufficient amount of NO making it the limiting reagent.
Question: For the balanced equation 2HNO3 + NO --> 3NO2 + H2O, if 98.1 grams of HNO3 is reacted with 18.4 grams of NO, how many grams of NO2 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 2HNO3 + NO --> 3NO2 +H2O, the ratio is going to be 2:1.
Step Two: Find the GFW of each compound (do not combine them).
HNO3: NO:
H: (1x1) + N: (1x14) + O: (3x16) = 63 GFW N: (1x14) + O: (1x16) = 30 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).
HNO3:
The number of grams for HNO3 is 98.1g. To find the number of moles you are going to divide 98.1g by 63 GFW. This equals 1.557 moles.
NO:
The number of grams for NO is 18.4g. To find the number of moles you are going to divide 18.4g by 30 GFW. This equals 0.613 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 2:1 HNO3 to NO.
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 2:1 ratio the number of moles of HNO3 should be double that of NO. Since you have 1.557 moles of HNO3 and 0.613 moles of NO, and you are supposed to have a 2:1 ratio, the number of moles for HNO3 should be 1.226 based on the number of moles of NO (0.613x2). Therefore, you have an excess of HNO3 meaning that you do not have the sufficient amount of NO 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 NO and the balanced equation is 2HNO3 + NO --> 3NO2 + H2O, the ratio between H2O and H2 is a 1:3 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 0.613 moles of NO (^number of moles of NO that we have^) we need triple the number of moles of NO2, as shown in the ratio, therefore, we multiply the number of moles of NO by 3 to get 1.839 moles of NO2.
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 NO2: N: (1x14) + O: (2x16)= 46 GFW grams=(1.839moles)(46GFW) ---> grams=84.594g
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Since the limiting reagent is NO and the balanced equation is 2HNO3 + NO --> 3NO2 + H2O, the ratio between H2O and H2 is a 1:3 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 0.613 moles of NO (^number of moles of NO that we have^) we need triple the number of moles of NO2, as shown in the ratio, therefore, we multiply the number of moles of NO by 3 to get 1.839 moles of NO2.
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 NO2: N: (1x14) + O: (2x16)= 46 GFW grams=(1.839moles)(46GFW) ---> grams=84.594g
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