Cl2 + 2NaBr ==> Br2 + 2NaCl
.x.....20g......16.0g...11.7g
The easy way to work the problem is to evoke the Law of Conservation of Mass. So x + 20 = 16 + 11.7 and solve for x. I think this is what the problem is illustrating; however, there is something wrong with the answer.
If we do it by stoichiometry, we can check to see if the 16 and 11.7 add up right.
mols Br2 = g/molar mass = 16.0/159.8 = about 0.1
mols NaCl is 2x that = 0.2
g NaCl = mols x molar mass = 0.2 x 58.44 = 11.7 and that checks out.
To check the Cl2 answer, mols Cl2 reacted = same as mols Br2 formed = 0.1.
Then g Cl2 = 0.1 mol x 70.9 = 7.09 which is not the same as the x above. (7.09 vs 7.7).
So which is the right answer? I would go with the last one since that agrees with the stoichiometry calculations. My guess is that the author of the problem overlooked this detail or you made a typo and the 20.0 for NaBr should have been 20.6.
when chlorine gas is bubbled into a solution of sodium bromide, the sodium bromide reacts to give bromine a red brown liquid and sodium chlorine. A solution was made by dissolving 20.0g of sodium bromide in 100.0 g of water. After passing through the solution, it contained 16.0 g of bromine and 11.7 g of sodium chloride. How many grams of chlorine reacted?
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