Mercury and bromine will react with each other to produce mercuric bromide. What mass of mercuric bromide can be produced from the reaction of 5.00 mL mercury (density 13.5 g/mL) and 5.00 mL bromine (density 3.2 g/mL)? What volume in mL of excess reactant remains?

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To solve this problem, we need to follow these steps:

1. Calculate the mass of mercury and bromine.
2. Determine the limiting reactant.
3. Calculate the mass of mercuric bromide produced from the limiting reactant.
4. Calculate the volume of excess reactant remaining.

Let's start solving the problem step by step:

1. Calculate the mass of mercury and bromine:
- Mass of mercury = volume of mercury x density of mercury
Mass of mercury = 5.00 mL x 13.5 g/mL
Mass of mercury = 67.5 g

- Mass of bromine = volume of bromine x density of bromine
Mass of bromine = 5.00 mL x 3.2 g/mL
Mass of bromine = 16.0 g

2. Determine the limiting reactant:
To determine the limiting reactant, we compare the molar ratios of mercury and bromine in the balanced equation. The balanced equation is not given, so we need to write it:

Hg + Br2 -> HgBr2

The balanced equation shows that the molar ratio between mercury (Hg) and bromine (Br2) is 1:1. Therefore, the reactant that has the smallest amount of substance (in moles) is the limiting reactant.

To calculate the number of moles of mercury and bromine, we need to convert their mass to moles using their molar masses.

- Molar mass of mercury (Hg) = 200.59 g/mol
- Molar mass of bromine (Br2) = 159.808 g/mol

Moles of mercury = mass of mercury / molar mass of mercury
= 67.5 g / 200.59 g/mol
≈ 0.336 mol

Moles of bromine = mass of bromine / molar mass of bromine
= 16.0 g / 159.808 g/mol
≈ 0.100 mol

Since the moles of bromine are smaller than the moles of mercury, bromine is the limiting reactant.

3. Calculate the mass of mercuric bromide produced from the limiting reactant:
The molar ratio between bromine and mercuric bromide is 1:1 from the balanced equation. Therefore, the mass of mercuric bromide can be calculated based on the moles of bromine present.

- Molar mass of mercuric bromide (HgBr2) = molar mass of mercury + 2 x molar mass of bromine
= 200.59 g/mol + 2 x 159.808 g/mol
= 520.206 g/mol

Mass of mercuric bromide = moles of bromine x molar mass of mercuric bromide
= 0.100 mol x 520.206 g/mol
= 52.021 g

So, the mass of mercuric bromide produced from the reaction is 52.021 g.

4. Calculate the volume of excess reactant remaining:
To calculate the volume of excess reactant remaining, we need to determine the number of moles of the excess reactant (mercury), and then convert it to the volume of the excess reactant.

- Moles of mercury reacted (from the balanced equation) = moles of bromine reacted = 0.100 mol

Moles of excess mercury = moles of mercury initially - moles of mercury reacted
= 0.336 mol - 0.100 mol
= 0.236 mol

To convert moles of mercury to volume (mL), we can use the molar volume at standard temperature and pressure (STP), which is 22.4 L/mol.

Volume of excess mercury = moles of excess mercury x molar volume at STP
= 0.236 mol x 22.4 L/mol x 1000 mL/L
= 5296 mL (rounded to the nearest whole number)

Therefore, approximately 5296 mL of excess mercury remains.

In summary:
- The mass of mercuric bromide produced from the reaction is 52.021 g.
- Approximately 5296 mL of excess mercury remains.