I really need help with finding the mass of NaOH in this problem

What mass of NaOH is needed to precipitate the Cd(2+) ions from 34.0 mL of 0.530 M Cd(NO3)2 solution?

I don't even know where to start.

2 NaOH + Cd(NO3)2 -----> 2 NaNO3 + Cd(OH)2

ok this is your equation ... and your precipitate is Cd(OH)2

number of moles of Cd(NO3)2 = Molarity x Volume (liters)

number of moles of Cd(NO3)2 = 0.53 M x 0.034 L = 0.018 moles

ok

from the equation

2 moles of NaOH ===> React with 1 mole of Cd(NO3)2

X moles of NaOH ===> react with 0.018 mole of Cd(NO3)2


X = 0.018 x 2 = 0.036 moles of NaOH


now mass of NaOH = moles x molar mass = 0.036 mole x 40 g/mole

mass of NaOH = 1.44 grams :D


hope you got it , you just need to get the Reaction correctly

2(Na+) + Cd(NO3)2 --> Cd(2+) + 2NaNO3

(34.0 mL Cd(NO3)2) x (0.530 moles Cd(NO3)2 / 1000 mL Cd(NO3)2)
x (2 moles Na+ / 1 mole Cd(NO3)2) x (1 mole NaOH / 1 mole Na+)
x (about 40g NaOH / 1 mole NaOH)

The units should all drop away (except grams :)) when you multiply the fractions