Question

To solve stoichiometry problems, you must always calculate numbers of moles. Recall that molarity, M, is equal to the concentration in moles per liter: M=mol/L.

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)

Part A

What mass of silver chloride can be produced from 1.39 L of a 0.224 M solution of silver nitrate?

Express your answer with the appropriate units.

mass of AgCl=

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Part B

The reaction described in Part A required 3.24 L of magnesium chloride. What is the concentration of this magnesium chloride solution?

Express your answer with the appropriate units.

Answer 1

A)

first, find mol o fAgNO3

mol = MV = 1.39*0.224 = 0.31136 mol of AgNO3

so..

if ratio s 2 mol of AGNO3 to 2 mol of AgCl3

then

1:1 ratio

so

0.31136 mol --> 0.31136 mol of AgCl

MW of AgCl = 143.32 g/mol

mass = mol*MW = 0.31136 *143.32 = 44.624 g of AgCl

b)

MgCl --> 3.24 L

find concnetration

so..

M = mol/V

we need mol of MGCl2 used

ratio was 2 mol of AgNO3 to 1 mol of MGCl2

0.31136 mol of AgNO3 --> 1/2*0.31136 = 0.15568 mol of MgCl2

so...

M = mol/V = 0.15568/(3.24) = 0.048049 M