Respuesta :
Answer:
a. Limiting reagent is the Hg.
b. 17.9 g of HgBr₂ will be produced in the reaction.
c. 6.5x10⁻³ moles of Br₂, are without reaction and the mass for that, is 1.03 g.
Explanation:
This is the reaction:
Hg (l) + Br₂ (l) --> HgBr₂ (s)
Ratio is 1:1, so 1 mol of mercury reacts with 1 mol of bromine.
Mass / Molar mass = Mol
10 g / 200.59 g/m = 0.0498 moles of Hg
9 g / 159.8 g/m = 0.0563 moles of Br₂
As ratio is 1:1, and for 0.0563 moles of Br₂ I need the same amount of Hg to react, I only have just 0.0498 so my limiting reactant is the Hg.
Ratio with products is also 1:1, so 1 mol of Hg makes 1 mol of bromide.
Then 0.0498 moles of Hg will produce 0.0498 moles of bromide.
Molar mass = 360.39 g/m
Mol . molar mass = mass → 0.0498 m . 360.39 g/m = 17.9 g of HgBr₂
To react all the Br₂ I have to consume the 0.0563 moles of it, but I only have 0.0498 moles for reacting, so the moles that will be left unreacted are:
0.0563 m - 0.0498 m = 6.5x10⁻³ moles of Br₂ (without reaction)
Mol . molar mass = mass → 6.5x10⁻³ mol . 159 g/m = 1.03 grams without reaction.
Answer:
a) The Hg is the limiting reactant
b) There is 17.98 grams of HgBr2 produced
c) There will rmain 0.0064 moles of Br2 unreacted, this is 1.02 grams
Explanation:
Step 1: Data given
Mass of Hg = 10.0 grams
Mass of Br2 = 9.00 grams
Molar mass of Hg = 200.59 g/mol
Molar mass of Br2 = 159.8 g/mol
Step 2: The balanced equation
Hg(l)+Br2(l) → HgBr2(s)
Step 3: Calculate moles Hg
Moles Hg =mass Hg / molar mass Hg
Moles Hg = 10.0 grams / 200.59 g/mol
Moles Hg = 0.0499 moles
Step 4: Calculate moles Br2
Moles Br2 = 9.00 grams / 159.8 g/mol
Moles Br2 = 0.0563 moles
Step 5: Calculate the limiting reactant
For 1 mol Hg we need 1 mol Br2 to produce 1 mol of HgBr2
Hg is the limiting reactant. It will completely be consumed (0.0499 moles)
Br2 is in excess. There will remain 0.0563 - 0.0499 = 0.0064 moles
This is 0.0064 moles * 159.8 = 1.02 grams
Step 6: Calculate moles HgBr2
For 1 mol Hg we need Br2 to produce 1 mol HgBr2
For 0.0499 moles Hg we'll have 0.0499 moles HgBr2
Step 7: Calculate mass of HgBr2
Mass HgBr2 = 0.0499 moles * 360.4 g/mol
Mass HgBr2 = 17.98 grams
