
Answer:
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Explanation:
Ā Ā Ā Ā 2AgNOā(aq) Ā Ā + Ā Ā NaāS(aq) => 2NaNOā(aq) + AgāS(s)
given Ā 2.0g Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā 2.5g Ā Ā Ā Ā Ā Spec Ions Ā Ā Ā Driving Force ppt
= 2.0g/169.8gĀ·molā»Ā¹ Ā Ā Ā Ā =2.5g/78gĀ·molā»Ā¹
= 0.012mol Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā =0.032mol
0.012/2=>0.006* Ā Ā Ā Ā Ā Ā Ā 0.032/1=>0.032*
Limiting Reactant => *Dividing moles by respective coefficient => Limiting Reactant is the smaller resulting value. Therefore AgNOā is the Limiting Reagent
Theoretical Yield of ppt AgāS(s) = 1/2(0.012 mol) = 0.006 mol AgāS(s) = 0.006mol(246gĀ·molā»Ā¹ ) = 1.48g AgāS(s)
%Yield = (Lab Yield/Theoretical Yield) x 100% = (1.25g/1.48g)100% = 84%
Reagent in Excess => NaāS(aq) => 0.032mol is given but only 0.016mol is consumed in the reaction. That is, (0.032 - 0.016)mol = 0.016mol Ā NaāS(aq) remains in excess = (0.016mol)(78gĀ·molā»Ā¹) = 1.25g NaāS(aq) in excess.