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If you react 65.5 grams of sodium and 102 grams of chlorine, how many grams of sodium...
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Let's look at the reaction of sodium and chlorine to make sodium chloride.
2Na + Cl2 --> 2NaCl
Sodium is a metal that exists as elemental Na. Chlorine is a gas that exists as the divalent molecule Cl2. Two moles of Na are required for each mole of Cl2 to produce 2 moles of NaCl salt. So we need to find which reagent is the limiting reagent to find how much NaCl will be produced. To find the limiting reagent, we need to convert both reagents from grams to moles. We do this by dividing by their molecular weights (which can be found through the periodic table).
65.5 g Na * (1 mole/22.99 g) = 2.849 moles Na
102 g Cl2 * (1 mole/70.9 g) = 1.439 moles Cl2
To fully react the 1.439 moles of Cl2, we would need twice that number of moles of Na (which would be 2.878 moles). Since we don't have quite that number of moles of Na available (close but not quite enough), that makes Na the limiting reagent. We have more than enough Cl2 to react all of the Na.
From the chemical reaction above we see that every 2 moles of Na produces 2 moles of NaCl. That is a 1 to 1 ratio. So as the limiting reagent, 2.849 moles of Na will produce 2.849 moles of NaCl. Now multiply by the molecular weight of NaCl to find the amount in grams.
2.849 moles NaCl * (58.44 g/mole) = 166.50 grams NaCl
So the final answer is 166.50 grams of NaCl
Posted by ncchemist on May 23, 2013 at 11:47 PM (Answer #1)
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