I think this question was just posted earlier, so I will use some of my earlier reply for this one.
(1) First, you need to know the molarity of the NaOH solution. If it is the same problem, it was made by dissolving 4 g of NaOH in 200 mL of water. If this is not how the NaOH was made, let me know.
molarity = moles / liters, and moles = grams / MW, so the molarity of the NaOH solution will be
4 g / 40.00 g/mol = 0.100 mol = 0.5 mol/L or 0.5 M
0.200 L 0.200 L
I'm just "plugging in" in order to calculate the moles and molarity. 
(2) Now you can calculate the grams of benzoic acid that will react with 20 mL of this solution (I think it is this solution!)
Since this is a 1:1 reaction, moles of NaOH = moles of benzoic acid.
moles of NaOH = volume in liters X molarity = 0.020 L X 0.5 mol/L = 0.01 mol of NaOH.
moles of benzoic acid = moles of NaOH = 0.01 mol.
grams of benzoic acid = moles X molar mass = 0.01 mol X 122.12 g/mol =
>> mass of benzoic = 122.12/0.02 = 6106 g <<
Errrr... Nope! Follow the steps above: molarity of NaOH (already done), then moles of NaOH in 20 mL of the solution (done), then moles of benzoic acid (same as the moles of NaOH in this reaction), and finally the grams of benzoic acid.
Be sure to write the units on all of your numbers. That helps to see that the units cancel correctly to give the right answer. Definitely useful once you get into the habit!
Steve
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