50.0mL of .10 silver nitrate is added to 50.0mL of .20M calcium chloride. A white precipitate forms. After the reaction is complete, calculate the amount of precipitate that formed.

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  1. 2AgNO3 + CaCl2 >>AgCl + Ca(NO3)2
    you started with .05moles of silver nitrate and
    .1 moles of calcium chloride
    So the reaction is limited by the small amount of silver nitrate.
    so the amount of silver chloride formed is 1/2*.05=.025 moles. Convert that to mass.
    check my thinking.

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