Question

When solutions of silver nitrate and magnesium chloride are mixed, silver chloride precipitates out of solution according to the equation 2AgNO₃(aq)+MgCl2(aq)→2AgCl(s)+Mg(NO₃)₂(aq).

What mass of silver chloride can be produced from 1.96 L of a 0.233 Msolution of silver nitrate? Express your answer with the appropriate units. Mass of AgCl=65.3

The reaction described in Part A required 3.46 L of magnesium chloride. What is the concentration of this magnesium chloride solution? Express your answer with the appropriate units.

Answer 1

2AgNO3(aq)+MgCl2(aq)→2AgCl(s)+Mg(NO3)2(aq)

no of moles of AgNO3 = Molarity x volume = 0.233 x 1.96 = 0.45668 moles

from the balanced equation it is clear that 2 moles of AgNO3 will give 2 moles of AgCl

so from 0.45668 moles of AgNO3 will produce 0.45668 moles of AgCl

mass of silver chloride = moles x molar mass = 0.45668 x 143.32 = 65.4 grams

Part B

from the balanced equation

2 moles of AgNO3 required one mole of MgCl2

0.45668 moles of AgNO3 required 0.45668/2 moles of MgCl2 = 0.22834 moles of MgCl2

now we have moles and voleme we can find the concentration

Molarity M = 0.22834 / 3.46 = 0.066M