When solutions of silver nitrate and magnesium chloride are mixed, silver chlori

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.

Explanation / Answer

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

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