Suppose that 26.7 mL of 0.490 M NaCl is added to 26.7 mL of 0.225 M AgNO3.?

Suppose that 26.7 mL of 0.490 M NaCl is added to 26.7 mL of 0.225 M AgNO3.

a) How many moles of AgCl would precipitate?

b) What are the concentrations in M of Ag+, Cl-, NO3- and Na+ in the reaction mixture after the reaction occurs?

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    Lv 7
    2 months ago
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    Moles of Cl- = 0.490 moles / L *0.0267 L = 0.01308 moles = moles Na+

    moles of Ag+ = 0.225 mole/L * 0.0267 L = 0.00601 moles = moles NO3-

    Ag++Cl- = AgCl

    according to the equation we have a 1:1 ratio of reactants

    we also see that Ag+ is the limiting reactant 

    so we will get  0.00601 moles of AgCl 

    the Ag+ will all precipitate so very little left in solution 

    moles of Cl- remaining =  0.01308 moles - 0.00601 moles = 0.00708 moles 

    these are in 2* 0.0267 L = 0.0534 L

    [Cl-] = 0.00708 moles / 0.0534 L = 0.132 M

    [NO3] = 0.00601 moles /  0.0534 L = 0.112 M

    [Cl-] = 0.01308 moles / 0.0534 L = 0.245 M 

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