If you react 5.00 mL Hg(l) (density = 13.6 g/mL) and 5.00 mL Br2(l) (density = 3.10 g/mL), what mass of which reactant is left unreacted?

Answer 1

we know that when Hg and Br₂ reacts, we get

Hg(l) + Br₂(l)→ HgBr₂(s)

where Mercury and Bromine react in a 1:1 ratio

molar mass of Hg (MM_Hg )= 200.59 g/mol

molar mass of Br₂ (MM_Br2 )= 159.808 g/mol

Given,

density of Hg (D_Hg )= 13.6 g/mL

density of Br₂ (D_Br2 )= 3.10 g/mL

5 mL Hg

5 mL Br₂

Let's calculate the number of moles of each Hg and Br_2, we get

moles of Hg= (5 mL Hg)(13.5 g Hg / 1mL Hg)(1 mol Hg / 200.59 g Hg)= 0.337 mol Hg

moles of Br₂ = (5 mL Br2)(3.10 g Br2 / 1mL Br2)(1 mol Br2 / 159.808 gBr2) = 0.097 mol Br₂

- Since they react in a 1:1 molar ratio, and there is more Mercury than Bromine, then Bromine is a limiting reagent and mercury is the excess reagent.

(0.337-0.097) moles = 0.24 moles of Hg remains unreacted

(0.24 moles of Hg) * (200.59 g of Hg/ 1 mol of Hg) = 48.14 g of Hg

Therefore, Hg is an excess reagent in the reaction so, 48.14 g of Hg remains unreacted.