how many moles of AgCl can be precipitated when excess NaCl is added to an equip

Question

how many moles of AgCl can be precipitated when excess NaCl is added to an equipped solution that contains 0.50 moles of silver nitrate? AgNo3(aq) + NaCl(aq) = AgCl(s)+ NaNO3(aq) how many moles of AgCl can be precipitated when excess NaCl is added to an equipped solution that contains 0.50 moles of silver nitrate? AgNo3(aq) + NaCl(aq) = AgCl(s)+ NaNO3(aq) how many moles of AgCl can be precipitated when excess NaCl is added to an equipped solution that contains 0.50 moles of silver nitrate? AgNo3(aq) + NaCl(aq) = AgCl(s)+ NaNO3(aq)

Explanation / Answer

here AgNO3 is the source for formation of AgCl we have only 0.5 moles of AgNO3 means only 0.5 moles ag we have.

so 0.5 moles of AgCl precipitate will form.

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