# Tutorial Question 5 Gives the problem and solution to tutorial question 5

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Published on 16 Oct 2011
School
University of Waterloo
Department
Nanotechnology Engineering
Course
NE121
Professor
Question 5 From Tutorial
A student gave you 1.20g of NaOH (s) and told you it contained Ca(OH)2.
precipitate.
What was the percentage by mas of NaOH in the original mixture.
Reaction 1:
2NaOH (s) + (Pb(NO3)2) (aq) 2NaNO3 (s) + Pb(OH)2 (aq)
Reaction 2:
Ca(OH)2 (aq) + Pb(NO3)2 (aq) Ca(NO3)2 (aq) + Pb(OH)2 (s)
Moles of product = Pb(OH)2 (s) made
=3.67g/ 241.216 g/mol = 0.01521 moles
= all moles of Ca(OH)2 reacted + half moles of NaOH reacted
0.01521 moles = ½ (moles of NaOH) + moles Ca(OH)2
Let x be the mass of NaOH reacted: This is what you are looking for to get % mass
0.1521 moles = ½ (x/39.998 g/mol) + (1.20 x)/ 74.094 g/mol
x= 0.9898g % by mass = 0.9898g/1.20g = 82.5 % of NaOH
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## Document Summary

A student gave you 1. 20g of naoh (s) and told you it contained ca(oh)2. To this you add lead nitrate solution (pb(no3)2) which produces 3. 67 g of lead hydroxide (pb(oh)2)) precipitate. What was the percentage by mas of naoh in the original mixture. 2naoh (s) + (pb(no3)2) (aq) 2nano3 (s) + pb(oh)2 (aq) Ca(oh)2 (aq) + pb(no3)2 (aq) ca(no3)2 (aq) + pb(oh)2 (s) = all moles of ca(oh)2 reacted + half moles of naoh reacted. 0. 01521 moles = (moles of naoh) + moles ca(oh)2. Let x be the mass of naoh reacted: this is what you are looking for to get % mass.