Consider the following aqueous reaction. Hg(NO₃) ₂ (aq) + Na₂S (aq) →HgS (s) + 2 NaNO₃ (aq) How many grams of HgS can be prepared from 50.0 mL of 0.0100 M Hg(NO₃) ₂ and 20.0 mL of 0.100 M Na₂S?
Molar Mass(HgS) = 232.7 g/mol

Question

Quizplus · Accepted Answer

First, we need to determine the limiting reagent in the reaction. To do this, we will use stoichiometry and the given concentrations of the reactants to determine how many moles of each reactant we have: - 50.0 mL of 0.0100 M Hg(NO₃)₂: (0.0100 mol/L) x (0.0500 L) = 0.000500 mol Hg(NO₃)₂ - 20.0 mL of 0.100 M Na₂S: (0.100 mol/L) x (0.0200 L) = 0.00200 mol Na₂S Now, we can use the stoichiometric coefficients from the balanced equation to determine how many moles of HgS can be produced from each reactant: - From Hg(NO₃)₂: 1 mol Hg(NO₃)₂ : 1 mol HgS So, 0.000500 mol Hg(NO₃)₂ will produce 0.000500 mol HgS. - From Na₂S: 1 mol Na₂S : 1 mol HgS So, 0.00200 mol Na₂S will produce 0.00200 mol HgS. Since the stoichiometric coefficient of Hg(NO₃)₂ is 1, while the stoichiometric coefficient of Na₂S is 2, we can see that we have less moles of the first reactant than we need for complete reaction. Therefore, Hg(NO₃)₂ is the limiting reagent. Now, we can use the moles of limiting reagent we have to determine how many moles of HgS we can produce: - From Hg(NO₃)₂: 0.000500 mol HgS Since the molar mass of HgS is 232.7 g/mol, we can convert the number of moles to grams: 0.000500 mol HgS x (232.7 g/mol) = 0.116 g HgS Therefore, the answer is D) 0.116 g HgS.

Consider the Following Aqueous Reaction