I realize this is a general chemistry stoichiometry question, however I can not fully grasp it.
I am supposed to calculate the theoretical amount (volume) of 3M sodium hydroxide (NaOH) needed to convert 1.5g of benzoic acid to its salt.
I got 0.012 moles of bezoic acid present at the beginning and by looking at the balanced equation I see that there is a 1:1 ratio between the benzoic acid and NaOH, however I have no idea where to go from here. I am so confused.
Could someone help as soon as possible? Thanks!
You've done the hard part.
Yes, 1.5 g benzoic acid is 0.123 moles which means you must have 0.123 moles NaOH. What's the definition of molarity?
M = #moles/L; rearrange that to
M x L = #moles
So, you have # moles and M, calculate L and round to two significant figures. (I know 0.123 is more than that but I routinely carry one more place than I need and round at the end; otherwise, sometimes I make a rounding error.)
Here is a worked example of a stoichiometry problem that you may want to copy. It's a good step by step procedure.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Thanks! and you do mean that it is 0.0123 moles, therefore I would plug in 3 for M, and 0.0123 for moles so that implies 3 x L = 0.0123 --> 0.0123/3 = L = .0041 L ~ 4.0mL ?
yes. that would be 4.1 mL of 3 M NaOH to add, theoretically, and if that is 3 M and not 3.0 M, then you may have only one s.f. and your 4 would be right (but not 4.0). If it is 3.0 M NaOH, I would round to 4.1 mL. Since benzoic acid can be titrated with NaOH, that really isn't a "highly" theoretical value but you may calculate exact amounts if the significant figures are there. I mean, in a quant lab, you could titrate to 4.12 mL or something like that. Of course, if it were a quant lab one wouldn't use 3 M NaOH to titrate it, either. :-).
Sure, I'd be happy to help you with this stoichiometry problem!
To calculate the theoretical amount (volume) of 3M sodium hydroxide (NaOH) needed to convert 1.5g of benzoic acid to its salt, we first need to use the molar mass of benzoic acid to calculate the number of moles present.
The molar mass of benzoic acid (C6H5COOH) is calculated as follows:
C: 6 * 12.01 = 72.06 g/mol
H: 5 * 1.008 = 5.04 g/mol
O: 2 * 16.00 = 32.00 g/mol
Total: 72.06 + 5.04 + 32.00 = 109.10 g/mol
So, to find the number of moles of benzoic acid, we divide the mass (1.5g) by the molar mass (109.10 g/mol):
moles = mass / molar mass = 1.5g / 109.10 g/mol = 0.0137 mol
Since the balanced equation shows a 1:1 ratio between benzoic acid and sodium hydroxide, this means that for every mole of benzoic acid, we will need one mole of sodium hydroxide. Therefore, we need 0.0137 mol of sodium hydroxide.
To find the volume of the 3M sodium hydroxide solution needed, we can use the definition of molarity, which is moles per liter (mol/L).
Molarity (M) = moles / volume (L)
We can rearrange this equation to solve for volume:
Volume (L) = moles / Molarity (M)
Plugging in the values, we have:
Volume (L) = 0.0137 mol / 3 mol/L = 0.0046 L, or 4.6 mL
Therefore, the theoretical amount (volume) of 3M sodium hydroxide needed to convert 1.5g of benzoic acid to its salt is 4.6 mL.
I hope this explanation helps you understand how to solve the problem! Let me know if you have any further questions.