Question

The metathesis reaction between sodium sulfide and chromium(IIII) chloride is shown below

3Na2S(aq)+2 CrCl3(aq) arrow Cr2S3(s)+6NaCl(aq)

How many grams of solid chromium (III) sulfide can be produced by the reaction of 250.0 ml of 0.500 M sodium sulfide solution with 275.0 ml of 0.250 M Chromium (III) chloride solution?

Answer 1

3 Na₂S (aq) + 2 CrCl₃ (aq) ---> Cr₂S₃ (s) + 6 NaCl (aq)

3 mole 2 mole 1 mole 6 mole

Given

Volume of sodium sulfide = 250 ml = 0.25 L

Molarity of sodium sulfide = 0.5 M = 0.5 mol/L

No. of moles of sodium sulfide = Volume * molarity = 0.25 L * 0.5 mol/L = 0.125 moles

Volume of Chromium chloride = 275 ml = 0.275 L

Molarity of Chromium chloride = 0.250 M or mol/L

No. of moles of Chromium chloride = 0.275 L * 0.25 mol/L = 0.06875 moles

according to reaction stoichometry 3 moles of sodium sulfide will require 2 moles of chromium chloride

so 0.06875 moles of chromium chloride requires 0.06875 moles * 3/2 = 0.1031 moles of sodium sulfide

we have sodium sulfide in excess

so 0.06875 moles of chroimum chloride will react

2 moles of chromium chloride will give 1 moles of chromium sulfide

so 0.06875 moles of chromium chloride will give 0.06875*1/2 = 0.034375 moles of chromium sulfide

Molar mass of chromium sulfide = 200.2 g/mol

Mass of chromium sulfide = no. of moles * molar mass = 0.034375 moles * 200.2 g/mol = 6.882 g

Answer is 6.882 g of chromium III sulfide will be formed