Question

If i start with 0.98g of NaOH how many moles of NaOH did we start with? I get 0.25 mol of NaOH, is this correct?

Answer 1

Using the equation \[n=(m/M)\] where M(NaOH) = (23.0+16.0+1.0)

So:
n(NaOH) = (0.98/40)
n(NaOH) = 0.0245

Answer 2

Okay, then if I want to convert that to moles of NaCl it would be .02 mol of NaCl which then would then convert to 1.17 g/mol of NaCl. Correct? (Just making sure my pre-lab is correct)

Answer 3

You've worded that wrong, the unit for mass (m) is just grams, the unit for molar mass (M) is g/mol
So 0.02 mole of NaCl with molar mass of (23.0+35.5) = 58.5 g/mol will have a mass of 1.17g

Answer 4

Thank you. Finally, if I started with 0.98g of NaOH and I had a final weight of 1.51 NaCl what is my percent yield? I then got this: (1.51NaCl(actual yield)/1.17(theoretical weight)x100%= 105%. I feel like I'm doing something wrong

Answer 5

Initial equation.....HCl + NaOH ---> NaCl + H2O

Answer 6

Yeah, just flip your fraction so it's
((Theoretical mass)/(Actual mass)) x 100 = %

So 100 * (1.17/1.51) = 77.483% = 77.5% (3.sf)