Question

please help!!

A 25.0 ml of 0.527 M sodium sulfide(aq) and 55.0 ml of 0.243 Miron (III) nitrate are mixed together. What mass of solid iron (II) sulfide can be formed? What is the limiting reagent? Show your work for full credit. 3 Na2S (aq) + 2 Fe(NO3)3 (aq) – Fe2S3 (s) + 6 NaNO3 (aq) The limiting reagent is Enter your answer here The theoretical yield of iron sulfide is Enter your answer here

Answer 1

*No.of moles = Molarity × Volume (L)

No.of moles of Na₂S = 0.527 × 0.025 = 0.013175

No.of moles of Fe(NO₃)₃ = 0.243 × 0.055 = 0.013365

    3 Na₂S + 2 Fe(NO₃)₃ -------> Fe₂S₃  + 6 NaNO₃

3 2 1

0.013175 0.013365 ---------------> moles taken

0.013175 (2/3) × 0.013175= 0.008783 -----> moles reacted

0 0.00458 -------------> moles left.

Since sodium sulphide is completely consumed, it is the limiting reagent.

*3 moles Na₂S produces -------> 1mole Fe₂S₃

0.013175 moles produces-------> ? = 0.013175/3

No.of moles of Na​​​​​​₂S  = 0.00439