CHEMISTRY !!!!!!!!!!!!!!!!!!!!!!!!?

: CHEMISTRY !!!!!!!!!!!!!!!!!!!!!!!!?
wastewater is treated with sodium sulfide solution in order to have a solid mercury (ll) sulfide and sodium nitrate solution result. in a lab, 50.0 mL of a 0.0100 M mercury (ll) nitrate are used. how many grams of sodium nitrate result?

Hg(NO3)2 + Na2S > HgS + NaNO3

Answers and Views:

Answer by Glavinon Chavello
moles of Hg(NO3)2 = Molarity * volume = 0.0100 * 50.0 = 0.500 mmoles

Hg(NO3)2 + Na2S > HgS + 2NaNO3

Sodium nitrate formed = 2/1 * 0.500 = 1mmoles = 0.001 moles

mass of NaNO3 = moles * Molar mass= 0.001 * 85 = 0.085 g