A 150-mL sample of a 0.600-mol/L calcium nitrate solution is mixed with 1.00 L o

Question

A 150-mL sample of a 0.600-mol/L calcium nitrate solution is mixed with 1.00 L of a 0.500-mol/L NaOH.  Calculate the mass of the precipitate produced in the reaction.

Explanation / Answer

First find out moles of Calcium Nitrate: .150 L x .600 mol/L= 0.09 moles Moles of NaOH: 0.500 moles Equation: Ca(NO3)2(aq) + NaOH(aq)~~~> Ca(OH)2 + Na(NO3)2(aq) Calcium is the limiting factor, so you'll end up with 0.09 moles ofCa(OH)2. The molecular mass is 74.093 grams per mole. 6.67 grams Technically a decent amount will be dissolved in solution, reducingthis mass. This can be figured out using Ksp The equation is Ksp=[OH]^2[Ca] The solution is basic enough and the Ksp is small enough that theamount in solution is negligible. It would reduce the totalmass to about 6.60 grams. I don't know where you are in thebook and if they expect you to calculate that or not, so I'm notgoing to do the exact calculations on that. You can do thatif necessary, all the equations are listed, and the Ksp and thingsare listed in your book.

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