Answer:
68.83 g HgS
Explanation:
The balanced reaction is:
Hg(C₂H₃O₂)₂ + Na₂S ---> 2 NaC₂H₃O₂ + HgS
Because Hg(C₂H₃O₂)₂ reacts completely, it is the limiting reactant. The solid precipitate is mercury (II) sulfide.
To find the amount of solid precipitate, you need to (1) convert grams Hg(C₂H₃O₂)₂ to moles Hg(C₂H₃O₂)₂ (via molar mass from periodic table), then (2) convert moles Hg(C₂H₃O₂)₂ to moles HgS (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles HgS to grams HgS (via molar mass from periodic table).
It is important to arrange the ratios/conversions in a way that allows for the cancellation of units. The desired unit should be the numerator. The final answer should have 4 sig figs because the given value with the lowest sig fig value (94.28 grams) has 4 sig figs.
Molar Mass (Hg(C₂H₃O₂)₂): 200.59 g/mol + 4(12.01 g/mol) + 6(1.008 g/mol) + 4(16.00 g/mol)
Molar Mass (Hg(C₂H₃O₂)₂): 318.678 g/mol
Molar Mass (HgS): 200.59 g/mol + 32.07 g/mol
Molar Mass (HgS): 232.66 g/mol
94.28 g Hg(C₂H₃O₂)₂ 1 mole 1 mole HgS 232.66 g
------------------------------- x ------------------ x ------------------------------ x ---------------- =
318.678 g 1 mole Hg(C₂H₃O₂)₂ 1 mole
= 68.83 g HgS
