Question

When solutions of silver nitrate and sodium chloride are mixed, silver chloride precipitates out of solution according to the equation AgNO3(aq)+NaCl(aq)→AgCl(s)+NaNO3(aq) 

A.) What mass of silver chloride can be produced from 1.73 L of a 0.278 M solution of silver nitrate? 

B.)The reaction described in Part A required 3.26 L of sodium chloride. What is the concentration of this sodium chloride solution

Answer 1

A.The given equation is

AgNO₃(aq)+NaCl(aq)→AgCl(s)+NaNO₃(aq)

From the above balanced equation stoichiometry, we see that 1 mole of silver nitrate precipitates 1 mole of silver chloride.

The reaction used 1.73 L of 0.278 M (0.278 mol L^-1) solution of silver nitrate.

No. of moles of AgNO₃ in 1.73 L of 0.278 mol L^-1 solution = 0.278 mol L^-1 * 1.73 L

= 0.48094 moles

As 1 mole of silver nitrate precipitates 1 mole of silver chloride, therefore 0.48094 moles of silver nitrate will precipitate 0.48094 moles of silver chloride.

Therefore,

No. of moles of AgCl precipitated = 0.48094 moles

Molar mass of AgCl = 143.32 g mol^-1

Mass of 0.48094 moles of AgCl = 143.32 g mol^-1 * 0.48094 moles

=68.93 g

= 68.9 g -------[rounding off to 3 significant figures]

Therefore , mass of AgCl precipitated = 68.9 g

B. From the balanced equation of reaction, we see that we see that 1 mole of AgNO₃ reacts completely with 1 mole of NaCl.

From part [A] we have, no. of moles of AgNO₃ reacted = 0.48094 moles

Therefore, no. of moles of NaCl that must have reacted = 0.48094 moles

Volume of NaCl solution used = 3.26 L

Concentration of NaCl solution used =  0.48094 moles / 3.26 L

= 0.150 mol L^-1

= 0.150 M