Question

In a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M mercury(II) nitrate. How many grams of mercury(II) sulfide (232.2 g/mol) form? Round answer to two sig figs and do not include units.

Answer 1

- Ans. The balanced equation is as: - Nazs caq.) + MG (NO3)2 (aq) 2NG NO3 (99.) + Hg S (s) Hg (NO3)2 = mer cury (11) nitrate Hgs = mercury (11) sulfide Nags= sodium sulfide given volume and concentration of Nas is = 0.02L1001M So, Concentration & male Volume moles of Na2S= 0.18 0.02 so, Moes of Na, S =0.002 and, Similarly moles of med cudy (11) Nitrate ax, moles of Hg(NO3)₂ = 0.0 5 xo.ol so, moles of My(NO3) = 0.0005

Shichiometric molar dation Mg(NO3)2: Nag S=1:1 actual molar ratio 0 Hg (NO3), : Na2S = .0005:6002 31:4 Therefore Nazs is present in excess and Mg (NO3)2 is limiting reagent » » So, 1 mode of Hg (NO3)2 produce Imole Hgs 0.0005 mole Hg (NO3)2 produce > 0.0005 mdle Hgs Now, mole = mass 19) molar mo So, mass 9 Hgs (g) = Moles x malad mass mass of HGS = 10.0005 x 232.2) mass of Hgs = 0.1161 gram/