Question

Step By Step please

In a simulation mercury removal from industrial wastewater, 0.050 L of 0.010 M mercury(II) nitrate reacts with 0.020 L of 0.10 M sodium sulfide. How many grams of mercury(II) sulfide form? Write a reaction table for this process.

Answer 1

Hg(NO3) (aq) + Na2S (aq) --------> HgS(s) + Na2NO3 (aq)

0.010 M Hg(NO3) (aq) react with   0.10 M Na2S (aq) to form products

multiplying both side by respective volumes

  0.010 mole / lit * 0.050 lit  Hg(NO3) react with 0.10 mole / lit * 0.020 lit Na2S (aq) to form products

5 * 10^-4  moles of   Hg(NO3) react with 2 * 10^-3     Na2S (aq)   

as in above we can clearly see that Hg(NO3) is the limiting reagent that means the product formed

HgS(s) = 5 * 10^-4  moles

As moles of HgS(s) = mass of HgS(s) / molar mass of HgS(s)

5 * 10^-4  moles * 232.66 g/mol = mass of HgS(s)

mass of HgS(s) = 0.1163 g