To solve stoichiometry problems, you must always calculate numbers of moles. Rec

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.

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.

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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.

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.

Explanation / Answer

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

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