Hg + Br2 -> HgBr2
1. What mass of HgBr2 can be produced from the reaction of 10.0 g Hg and 9.00 g Br2? What mass of which reagent is left unreacted?
2. What mass of HgBr2 can be produced from the reaction of 5.00 mL mercury (density = 13.6 g/mL) and 5.00 mL bromine (density = 3.10 g/mL)?
See the procedure for CaSO4 and H3PO4 above.
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To solve these problems, we need to determine the limiting reagent, calculate the amount of product formed from the limiting reagent, and then calculate the excess reagent remaining.
1. First, we need to determine the limiting reagent. To do this, we compare the number of moles of each reactant to the stoichiometry of the balanced equation.
a. Convert the mass of each reactant to moles:
Moles of Hg = 10.0 g Hg / molar mass of Hg
Moles of Br2 = 9.00 g Br2 / molar mass of Br2
b. Use the stoichiometry of the balanced equation to determine the mole ratio of Hg to Br2 in the reaction.
From the balanced equation: 1 mol Hg reacts with 1 mol Br2 to produce 1 mol HgBr2.
c. Compare the moles of Hg and Br2 to determine the limiting reagent. The limiting reagent is the one that produces the lesser amount of product.
Let's assume that the molar mass of Hg is 200.59 g/mol and the molar mass of Br2 is 159.808 g/mol.
2. Now, let's calculate the amount of HgBr2 produced from the limiting reagent.
a. Determine the moles of Hg that react:
Moles of Hg = moles of Hg (from step 1a)
The molar ratio tells us that 1 mol of Hg reacts with 1 mol of Br2.
b. Calculate the moles of HgBr2 produced using the mole ratio:
Moles of HgBr2 = Moles of Hg
c. Convert the moles of HgBr2 to grams using the molar mass of HgBr2:
Mass of HgBr2 = Moles of HgBr2 * molar mass of HgBr2
3. Finally, let's calculate the mass of the unreacted reagent.
a. Determine the moles of Br2 that react:
Moles of Br2 = moles of Br2 (from step 1a)
b. Calculate the moles of HgBr2 formed from the moles of Br2 using the mole ratio:
Moles of HgBr2 = Moles of Br2
c. Calculate the mass of the excess reagent remaining:
Mass of excess Br2 = Moles of Br2 * molar mass of Br2
Now, let's solve the problem step-by-step:
1. What mass of HgBr2 can be produced from the reaction of 10.0 g Hg and 9.00 g Br2? What mass of which reagent is left unreacted?
Step 1: Determine the limiting reagent
a. Moles of Hg = 10.0 g Hg / 200.59 g/mol = 0.04982 mol Hg
b. Moles of Br2 = 9.00 g Br2 / 159.808 g/mol = 0.05629 mol Br2
Step 2: Calculate the amount of HgBr2 produced from the limiting reagent
a. Moles of HgBr2 = Moles of Hg = 0.04982 mol Hg
b. Mass of HgBr2 = Moles of HgBr2 * molar mass of HgBr2
Mass of HgBr2 = 0.04982 mol * (200.59 g/mol + 159.808 g/mol)
Mass of HgBr2 = 20.0015 g HgBr2
Step 3: Calculate the mass of the unreacted reagent
a. Moles of excess Br2 = Moles of Br2 - Moles of HgBr2
Moles of excess Br2 = 0.05629 mol - 0.04982 mol = 0.00647 mol
b. Mass of excess Br2 = Moles of excess Br2 * molar mass of Br2
Mass of excess Br2 = 0.00647 mol * 159.808 g/mol = 1.0326 g Br2
Therefore, the mass of HgBr2 produced is 20.0015 g and the mass of the unreacted Br2 is 1.0326 g.
2. What mass of HgBr2 can be produced from the reaction of 5.00 mL mercury (density = 13.6 g/mL) and 5.00 mL bromine (density = 3.10 g/mL)?
Step 1: Determine the limiting reagent
a. Convert the volume of each reactant to mass:
Mass of Hg = volume of Hg * density of Hg
Mass of Hg = 5.00 mL * 13.6 g/mL = 68.0 g Hg
Mass of Br2 = 5.00 mL * 3.10 g/mL = 15.5 g Br2
b. Calculate the moles of each reactant:
Moles of Hg = mass of Hg / molar mass of Hg
Moles of Br2 = mass of Br2 / molar mass of Br2
Step 2: Calculate the amount of HgBr2 produced from the limiting reagent
a. Determine the moles of Hg that react:
Moles of Hg = mass of Hg / molar mass of Hg
b. Calculate the moles of HgBr2 using the mole ratio:
Moles of HgBr2 = Moles of Hg
c. Convert the moles of HgBr2 to grams using the molar mass of HgBr2:
Mass of HgBr2 = Moles of HgBr2 * molar mass of HgBr2
Step 3: Calculate the mass of the unreacted reagent
a. Determine the moles of Br2 that react:
Moles of Br2 = mass of Br2 / molar mass of Br2
b. Calculate the moles of HgBr2 formed from the moles of Br2 using the mole ratio:
Moles of HgBr2 = Moles of Br2
c. Calculate the mass of the excess reagent remaining:
Mass of excess Br2 = Moles of Br2 * molar mass of Br2
Now, plug in the values given and solve for the mass of HgBr2 and the mass of the unreacted reagent:
Step 1: Determine the limiting reagent
a. Mass of Hg = 68.0 g Hg
Mass of Br2 = 15.5 g Br2
Step 2: Calculate the amount of HgBr2 produced from the limiting reagent
a. Moles of Hg = 68.0 g Hg / molar mass of Hg = 68.0 g / 200.59 g/mol = 0.3397 mol Hg
b. Moles of HgBr2 = Moles of Hg = 0.3397 mol Hg
c. Mass of HgBr2 = Moles of HgBr2 * molar mass of HgBr2
Mass of HgBr2 = 0.3397 mol * (200.59 g/mol + 159.808 g/mol)
Mass of HgBr2 = 109.107 g HgBr2
Step 3: Calculate the mass of the unreacted reagent
a. Moles of Br2 = 15.5 g Br2 / molar mass of Br2 = 15.5 g / 159.808 g/mol = 0.0968 mol Br2
b. Moles of HgBr2 = Moles of Br2 = 0.0968 mol
c. Mass of excess Br2 = Moles of excess Br2 * molar mass of Br2
Mass of excess Br2 = 0.0968 mol * 159.808 g/mol = 15.497 g Br2
Therefore, the mass of HgBr2 produced is 109.107 g and the mass of the unreacted Br2 is 15.497 g.
To determine the mass of HgBr2 produced and the mass of each reagent that is left unreacted in the given chemical equation, we need to follow a series of steps.
1. What mass of HgBr2 can be produced from the reaction of 10.0 g Hg and 9.00 g Br2?
Step 1: Convert the given masses of Hg and Br2 into moles using their molar masses.
- The molar mass of Hg is 200.59 g/mol.
- The molar mass of Br2 is 159.80 g/mol.
The number of moles of Hg = Mass of Hg / Molar mass of Hg
The number of moles of Br2 = Mass of Br2 / Molar mass of Br2
Step 2: Determine the limiting reagent.
The limiting reagent is the one that limits the amount of product formed. To find the limiting reagent, we compare the moles of each reagent to their stoichiometric coefficients in the balanced equation (1:1 ratio for Hg and Br2).
Step 3: Calculate the moles of HgBr2 produced.
Since the balanced equation shows a 1:1 stoichiometric ratio between Hg and HgBr2, the number of moles of HgBr2 produced will be equal to the number of moles of the limiting reagent.
Step 4: Convert the moles of HgBr2 produced into grams using its molar mass.
- The molar mass of HgBr2 is 360.39 g/mol.
The mass of HgBr2 produced = Moles of HgBr2 produced * Molar mass of HgBr2
Step 5: Calculate the mass of the unreacted reagent.
To find the mass of the unreacted reagent, we subtract the mass of the reacted reagent from the initial mass of that reagent.
2. What mass of HgBr2 can be produced from the reaction of 5.00 mL mercury (density = 13.6 g/mL) and 5.00 mL bromine (density = 3.10 g/mL)?
Step 1: Convert the given volumes of Hg and Br2 into masses using their densities.
- The mass of Hg = Volume of Hg * Density of Hg
- The mass of Br2 = Volume of Br2 * Density of Br2
Step 2: Repeat steps 2 to 5 from the first question to determine the mass of HgBr2 produced and the mass of the unreacted reagent.
By following these steps, you can solve both questions and find the desired masses.