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.

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 1.27 L of a 0.192 M solution of silver nitrate?

Express your answer with the appropriate units.

Part B

The reaction described in Part A required 3.09 L of sodium chloride. What is the concentration of this sodium chloride solution?

Express your answer with the appropriate units.

Explanation / Answer

mass of AgCl

from V = 1.27 L M = 0.192 M of AgNO3

calculate moles of AgNO3 = M* = 1.27*0.192 = 0.24384 mol of AgNO3

then we can precipitate as many as

0.24384 mol of Ag+

or

0.24384 AgCl

MW AgCl = 143.32

mass = mol*MW = 0.24384 *143.32 = 34.94714 g of AgCl

B)

V = 3.09 of NaCl

find concnetration of NaCl

M = mol /V

ratio is 1mol of NaCl : 1 mol of Cl- : 1 mol of AGcl

therefore

mol = 0.24384

M = mol/V = 0.24384 /3.09 = 0.07891 M of NaCl

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