Question

a 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) Part A What mass of silver chloride can be produced from 1.51 L of a 0.293 M solution of silver nitrate? Express your answer with the appropriate units.

b. The reaction described in Part A required 3.54 L of sodium chloride. What is the concentration of this sodium chloride solution? Express your answer with the appropriate units.

Answer 1

Part A To find the mass of silver chloride Given that 151 L of a 0.293 M solution of silver nitrate AgNO3(aq) + NaCl(aq) → AgCl (s) + NaNO3(aq) First find the moles of silver nitrate moles =molarity x volume = 0.293 mol/Lx151 L = 0.442moles Find the moles of silver chloride According to equation 1 moles of AgNO3 gives 1 mole of AgCl Therefore moles of AgCl = 0.442moles To find the mass of silver chloride mass = moles Xmolarmass molarmass of silver chloride= 143.32g/mol mass = 0.442m ol x143.32g/mol = 63.35g Hence the mass of silver chloride= 63.35g Part B To find the concentration of sodium chloride solution Given that 3.54L of sodium chloride AgNO3(aq) + NaCl(aq) → AgCl (s)+NaNO3(aq) First find the moles of sodium chloride According to equation 1 moles of AgNO, reacts with 1 mole of NaCl Therefore moles of NaCl = 0.442 moles To find the concentration (molarity) Molarity - moles liters 0.442moles 3.54L = 0.125M Hence the concentration of sodium chloride solution = 0.125M|