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?

This is a limiting reagent (LR) problem and I know this because amounts are given for BOTH reactants. Therefore, one will be used completely (the LR) and the other will have some un-reacted.

Hg + Br2 ==> HgBr2

Use density to convert mL Hg to grams.
Do the same for mL Br2 to grams.

Convert g Hg to mols. mol = grams/atomic mass
Convert g Brw to mols. mol = g/molar mass.

Using the coefficients in the balanced equation convert mols Hg to mol HgBr.
Do the same for mols Br2 to mols HgBr2.
It is likely that these two values will not agree; the correrct answer in LR problems is ALWAYS the smaller value and the reagent producing that number is the LR..

Using the smaller value, convert to grams.
grams = mols x molar mass

To calculate the mass of mercuric bromide produced, we first need to determine the number of moles of each reactant. Then we can use stoichiometry to find the number of moles of mercuric bromide produced. Finally, we can convert the moles of mercuric bromide to grams using its molar mass.

Step 1: Calculate the number of moles of mercury.
Given: Volume of mercury = 5.00 mL and density of mercury = 13.5 g/mL
To find the mass of mercury, we multiply its volume by its density:
Mass of mercury = 5.00 mL * 13.5 g/mL = 67.5 g

To convert the mass of mercury to moles, we divide it by its molar mass. The molar mass of mercury is 200.59 g/mol.
Number of moles of mercury = 67.5 g / 200.59 g/mol = 0.3365 mol

Step 2: Calculate the number of moles of bromine.
Given: Volume of bromine = 5.00 mL and density of bromine = 3.2 g/mL
To find the mass of bromine, we multiply its volume by its density:
Mass of bromine = 5.00 mL * 3.2 g/mL = 16.0 g

To convert the mass of bromine to moles, we divide it by its molar mass. The molar mass of bromine is 79.90 g/mol.
Number of moles of bromine = 16.0 g / 79.90 g/mol = 0.2002 mol

Step 3: Determine the limiting reactant.
To determine the limiting reactant, we compare the number of moles of each reactant. The reactant with the smaller number of moles is the limiting reactant. In this case, bromine has 0.2002 moles, while mercury has 0.3365 moles. Therefore, bromine is the limiting reactant.

Step 4: Use stoichiometry to calculate the moles of mercuric bromide produced.
From the balanced chemical equation, we know that the molar ratio between mercury and mercuric bromide is 1:1. Therefore, the moles of mercuric bromide produced is also 0.2002 mol.

Step 5: Convert the moles of mercuric bromide to grams.
To calculate the mass of mercuric bromide, we multiply its moles by its molar mass. The molar mass of mercuric bromide (HgBr2) is 327.15 g/mol.
Mass of mercuric bromide = 0.2002 mol * 327.15 g/mol = 65.47 g

Therefore, the mass of mercuric bromide produced is 65.47 g.

To find the volume of excess reactant remaining, we need to determine the moles of bromine that reacted. Since bromine is the limiting reactant, all of it is consumed in the reaction.

Step 6: Calculate the volume of bromine that reacted.
To calculate the volume, we use the volume of bromine and its density.
Volume of bromine reacted = 5.00 mL

Since bromine is completely consumed, the volume of excess reactant remaining is zero.

Therefore, the volume of excess reactant remaining is 0.00 mL.

To determine the mass of mercuric bromide produced, we need to calculate the amount of each reactant, and then determine which one is the limiting reactant.

First, we calculate the mass of mercury:
Mass of mercury = volume of mercury x density of mercury
Mass of mercury = 5.00 mL x 13.5 g/mL

Next, we calculate the mass of bromine:
Mass of bromine = volume of bromine x density of bromine
Mass of bromine = 5.00 mL x 3.2 g/mL

To find the limiting reactant, we need to compare the moles of each reactant. We can convert the mass of each reactant to moles using their respective molar masses.

The molar mass of mercury (Hg) is 200.6 g/mol, and the molar mass of bromine (Br2) is 159.8 g/mol.

Moles of mercury = mass of mercury / molar mass of mercury
Moles of bromine = mass of bromine / molar mass of bromine

Next, we need to compare the mole ratio between mercury and bromine in the balanced chemical equation for the reaction. The balanced equation is:

Hg + Br2 -> HgBr2

The mole ratio is 1:1, meaning one mole of mercury reacts with one mole of bromine to produce one mole of mercuric bromide.

To determine the limiting reactant, we compare the number of moles of each reactant. The reactant that produces fewer moles of product is the limiting reactant.

Now, calculate the limiting reactant:
Limiting reactant = min(moles of mercury / 1,mole of bromine / 1)

Once we identify the limiting reactant, we can use the mole ratio from the balanced equation to calculate the moles of mercuric bromide produced.

Moles of mercuric bromide produced = moles of limiting reactant

Finally, we can convert the moles of mercuric bromide to its mass using its molar mass:

Mass of mercuric bromide = moles of mercuric bromide x molar mass of mercuric bromide

To find the volume in mL of the excess reactant that remains, we need to determine how many moles of excess reactant were not consumed during the reaction.

Moles of excess reactant = moles of excess reactant - (moles of limiting reactant x 1)

Volume of excess reactant = moles of excess reactant / density of excess reactant

By following these steps, you should be able to calculate the mass of mercuric bromide produced and the volume in mL of the excess reactant that remains.