When 2.00 g of silver nitrate and 2.50 g of sodium sulfide in different solutions combine to react, 1.25 g of precipitate are collected. * The text box will not allow you to use subscript. If you want to type the chemical formula for iron(III) sulfate, Fe2(SO4)3, it will need to be typed as Fe2(SO4)3. Remember that the symbol for iron, Fe, is not "fe" or "FE." Do not include the state of matter for each formula. • What is the chemical formula* for the product? Remember that spectator ions do not form a product in solutions. • What is the theoretical yield, in gram(s) for the solid product? • What is the chemical formula for the limiting reagent? • What is the chemical formula for the excess reagent? • What is the percent yield for the reaction? • How many gram(s) of the excess reagent is/are left over?

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Explanation:

        2AgNO₃(aq)    +    Na₂S(aq) => 2NaNO₃(aq) + Ag₂S(s)

given  2.0g                        2.5g          Spec Ions       Driving Force ppt

= 2.0g/169.8g·mol⁻¹        =2.5g/78g·mol⁻¹

= 0.012mol                      =0.032mol

0.012/2=>0.006*              0.032/1=>0.032*

Limiting Reactant => *Dividing moles by respective coefficient => Limiting Reactant is the smaller resulting value. Therefore AgNO₃ is the Limiting Reagent

Theoretical Yield of ppt Ag₂S(s) = 1/2(0.012 mol) = 0.006 mol Ag₂S(s) = 0.006mol(246g·mol⁻¹ ) = 1.48g Ag₂S(s)

%Yield = (Lab Yield/Theoretical Yield) x 100% = (1.25g/1.48g)100% = 84%

Reagent in Excess => Na₂S(aq) => 0.032mol is given but only 0.016mol is consumed in the reaction. That is, (0.032 - 0.016)mol = 0.016mol  Na₂S(aq) remains in excess = (0.016mol)(78g·mol⁻¹) = 1.25g Na₂S(aq) in excess.

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