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Question: For the balanced equation 2Na + 2H2O --> 2NaOH + H2, what would be the limiting reagent if 61.0 grams of Na were reacted with 26.9 grams of H2O?
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is 2Na + 2H2O --> 2NaOH + H2, the ratio is going to be 2:2.
Step Two: Find the GFW of each compound (do not combine them).
Na: H2O:
Na: 1x23 = 23 GFW H: (2x1 = 2) + O: (1x16 = 16) = 18 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).
Na:
The number of grams for Na is 61.0g. To find the number of moles you are going to divide 61.0g by 23 GFW. This equals 2.65 moles.
H2O:
The number of grams for H2O is 26.9g. To find the number of moles you are going to divide 26.9g by 18 GFW. This equals 1.49 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 2:2 ratio, we can divide both sides by 2 to make it a 1:1 ratio, Na to H2O.
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 1:1 ratio the number of moles of H2O and Na should be the same. Since you have 1.49 moles of H2O and 2.65 moles of Na, you do not have the sufficient amount of H2O makes it the limiting reagent.
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Question: For the balanced equation 2Na + 2H2O --> 2NaOH + H2, what would be the limiting reagent if 61.0 grams of Na were reacted with 26.9 grams of H2O?
Step One: Based on the coefficients of the reactants (the two compounds) find the ratio.
Since the balanced equation is 2Na + 2H2O --> 2NaOH + H2, the ratio is going to be 2:2.
Step Two: Find the GFW of each compound (do not combine them).
Na: H2O:
Na: 1x23 = 23 GFW H: (2x1 = 2) + O: (1x16 = 16) = 18 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).
Na:
The number of grams for Na is 61.0g. To find the number of moles you are going to divide 61.0g by 23 GFW. This equals 2.65 moles.
H2O:
The number of grams for H2O is 26.9g. To find the number of moles you are going to divide 26.9g by 18 GFW. This equals 1.49 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 2:2 ratio, we can divide both sides by 2 to make it a 1:1 ratio, Na to H2O.
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 1:1 ratio the number of moles of H2O and Na should be the same. Since you have 1.49 moles of H2O and 2.65 moles of Na, you do not have the sufficient amount of H2O makes it the limiting reagent.
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