Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

Help with Solving Limiting Reactant Problem?

The equation for the reaction is:

Ca(OH)2 (aq) + 2HCl (aq) ---> CaCl2 (aq) + 2H2O (l)

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  • 1 month ago
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    (29.4 g Ca(OH)2) / (74.0927 g Ca(OH)2/mol) = 0.39680 mol Ca(OH)2

    (32.0 g HCl) / (36.4611 g HCl/mol) = 0.87765 mol HCl

    0.39680 mole of Ca(OH)2 would react completely with 0.39680 x (2/1) = 0.79360 mole of HCl, but there is more HCl present than that, so HCl is in excess and Ca(OH)2 is the limiting reagent.

    (0.39680 mol Ca(OH)2) x (1 mol CaCl2 / 1 mol Ca(OH)2) x (110.984 g CaCl2/mol) =

    44.0 g CaCl2

    ((0.87765 mol HCl initially) - (0.79360 mol HCl reacted)) x (36.4611 g HCl/mol) =

    3.06 g HCl left over

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