Question

FeCl3(aq) + K2S(aq) → Fe2S3(aq) + KCl(aq) Consider the unbalanced equation above. What is the maximum mass of Fe2S3 that can be produced when 39.0 mL of 0.200 M FeCl3 and 37.0 mL of 0.190 M KZS react?

Answer 1

KCI S Balanced chemical equation: 2 FeCl3 + 3 K2S →1 Fe2S3 + 6 Theoretical moles ratio of FeCl3 : K2S Calculating number of moles: volume of FeCl3 = 39.0 mL = 0.0390 L Molarity of FeCl3 = 0.2 M Moles = Molarity Volume Moles of FeCl3 =( 0.2 0.0390 ) Moles of FeCl3 = 0.00780 mol 0.037 L volume of K2S = 37.0 mL = Molarity of K2S = 0.19 M Moles = Molarity Volume Moles of K2S =( 0.19 Moles of K2S = 0.00703 mol 0.037 ) Experimental moles ratio of Fec|3 : K2S = 0.00780 : 0.00703 1.110 : 1.000 3.329 : 3.000

Limiting reactant is K2S Excess reactant is FeCl3 Mass of products depends upon limiting reactant From balanced equation: 3 moles of K2S produces 1 0.00703 moles of K2S produces ? moles of Fe2s3 = ( 1 / 3 moles of Fe253 = 0.00234 mol Calculating mass: Mass = Moles x Molar mass mass of Fe253 = 0.002 x 207.9 mass of Fe2s3 = 0.4872 g Theoretical Yield of Fe2s3 = 0.487 g moles of Fe253 moles of Fe253 ) * 0.00703