1a) Iron (II) chloride and sodium sulfide undergo adouble-replacement reaction w

Question

1a) Iron (II) chloride and sodium sulfide undergo adouble-replacement reaction when mixed. Write a balanced chemicalequation for the reaction.
1b) If 40.0 g of iron (II) chloride and 40.0 g of sodium sulfideare mixed, what mass of iron (II) sulfide is produced? 1c) What is the limiting reactant? 1d) If 7.90 g of iron (II) sulfide is actually recovered, whatis the percent yield for the reaction? 1a) Iron (II) chloride and sodium sulfide undergo adouble-replacement reaction when mixed. Write a balanced chemicalequation for the reaction.
1b) If 40.0 g of iron (II) chloride and 40.0 g of sodium sulfideare mixed, what mass of iron (II) sulfide is produced? 1c) What is the limiting reactant? 1d) If 7.90 g of iron (II) sulfide is actually recovered, whatis the percent yield for the reaction?

Explanation / Answer

FeCl2 + Na2S -> 2NaCl + FeS n(FeCl2) = m/M = 40/126.75 = 0.3156mol n(Na2S) = m/M = 40/55.05 = 0.7266mol Stoichiometry ratio = n(FeCl2)/n(Na2S) = 1/1= 1 Actual ratio = 0.3156/0.7266 = 0.4344 Stoichiometry ratio > Actual Ratio Thus, FeCl2 is the limiting reagent n(FeS) = 1/1 x n(FeCl2) = 0.3156mol m(FeS) = nM = 0.3156 x 87.91 = 27.74g percent yield = 7.9/27.74 x 100 = 28.48% Hope this helps!

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