Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

35.0 g of BaCl2 is added to 1.0 L of 0.65 M of AgNO3. The following reaction occurs: 2 AgNO3(aq)+BaCl2(aq)→2 AgCl(s)+Ba(NO3)2(aq)?

Calculate the mass of the precipitate, AgCl, that will be produced and the moles of the reactant in excess that will remain in the solution after the reaction is complete.

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  • 1 month ago

    2 AgNO3 + BaCl2 → 2 AgCl + Ba(NO3)2

    (35.0 g BaCl2) / (208.233 g BaCl2/mol) = 0.16808 mol BaCl2

    (1.0 L) x (0.65 mol AgNO3/L) = 0.65 mol AgNO3

    0.16808 mole of BaCl2 would react completely with 0.16808 x (2/1) = 0.33616 mole of AgNO3, but there is more AgNO3 present than that, so AgNO3 is in excess and BaCl2 is the limiting reactant.

    (0.16808 mol BaCl2) x (2 mol AgCl / 1 mol BaCl2) x (143.3212 g AgCl/mol) =

    48.2 g AgCl produced

    (0.65 mol AgNO3 initially) - (0.33616 mol AgNO3 reacted) =

    0.31 mol AgNO3 left over

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