Asolution of nacl(aq)nacl(aq) is added slowly to a solution of lead nitrate, pb(no3)2(aq)pb(no3)2(aq) , until no further precipitation occurs. the precipitate is collected by filtration, dried, and weighed. a total of 13.25 g pbcl2(s)13.25 g pbcl2(s) is obtained from 200.0 ml200.0 ml of the original solution. calculate the molarity of the pb(no3)2(aq)pb(no3)2(aq) solution.

The molarity of the Pb(NO3)2 solution is 0.24 M

Explanation:

Assuming there was no loss during filtration, we can calculated amount of moles Pb(NO3)2 from the 13.25 g of PbCl2 ussing molar mass of  PbCl2

13.25 g / 278.1 g = 0.04 moles of PbCl2

According balanced equation 2NaCl + Pb(NO3)2 = 2NaNO3 + PbCl2, 1 mol of Pb(NO3)2 produces 1 mol of PbCl2. So, 0.04 moles of Pb(NO3)2 produces PbCl2

Molarity is = moles of solute/volumen of solution (L)=0.04/0.2= 0.24

I hope my answer helps you

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g