Mercury and bormine react with each other to produce mercury(II)bromide:

Hg + Br2 -> HgBr2

a) What mass of HgBr2 can be produced from the reactiion 10g HG and 9g Br2? [the part that really confuses me] What mass of which reagent is left unreacted?

b) What is the mass of HgBr2 can be produced form the reaction 5ml HG (density=13.6) and 5ml bromine (density=3.1)?

This is the question. I cannot get the right answer, and I'm unsure on how to do it. Can you pleaseeeeeeeeeeeeeee help me?

You get around the "which reagent is in excess" by the following.

Convert 10 g Hg t moles. moles = grams/molar mass.
Convert 9 g Br2 to moles the same way.

Now, using the coefficients in the balanced equation, convert moles Hg to moles of the product.
Do the same for the moles Br2 to moles of the product.
It is likely that the two answers will not agree which means one of them is wrong. The correct answer, in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. This should take care of your problem.
Post your work if you get stuck.

Of course, I'd be happy to help you understand how to solve these problems!

a) To determine the mass of HgBr2 produced and the mass of the unreacted reagent, we need to find the limiting reagent in the reaction. The limiting reagent is the one that is completely consumed in the reaction, determining the maximum amount of product that can be formed. We can compare the amount of each reactant to their respective stoichiometric coefficients in the balanced equation to find the limiting reagent.

1. Convert the mass of Hg to moles:
Mass of Hg = 10g
Molar mass of Hg = 200.59 g/mol (from periodic table)
Moles of Hg = Mass of Hg / Molar mass of Hg

2. Convert the mass of Br2 to moles:
Mass of Br2 = 9g
Molar mass of Br2 = 159.81 g/mol (from periodic table)
Moles of Br2 = Mass of Br2 / Molar mass of Br2

3. We need to compare the mole ratios of Hg and Br2 in the balanced equation:
Balanced equation: Hg + Br2 -> HgBr2
From the equation, we see that the ratio of Hg to Br2 is 1:1.

4. Determine the limiting reagent:
Compare the mole ratio of Hg to Br2 with the actual mole ratio:
Moles of Hg / Moles of Br2
If the mole ratio is greater than 1, Hg is in excess and Br2 is the limiting reagent. If the mole ratio is less than 1, Hg is the limiting reagent.

5. Calculate the moles of HgBr2 formed based on the limiting reagent:
Since Br2 is the limiting reagent, we will use the moles of Br2 to determine the moles of HgBr2 formed. Moles of HgBr2 = Moles of Br2 (based on the mole ratio from the balanced equation)

6. Calculate the mass of HgBr2 formed:
Mass of HgBr2 = Moles of HgBr2 * Molar mass of HgBr2

7. Calculate the remaining 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.

b) To answer this question, we need to use the volume and density of the liquids given. Remember, density is calculated by dividing mass by volume (density = mass/volume). Follow the same steps as in part a) to find the limiting reagent and calculate the mass of HgBr2 formed.

1. Convert the volumes of Hg and Br2 to mass using their respective densities:
Mass of Hg = Volume of Hg * Density of Hg
Mass of Br2 = Volume of Br2 * Density of Br2

2. Convert the mass of Hg and Br2 to moles using their respective molar masses.

3. Compare the mole ratios to find the limiting reagent.

4. Calculate the moles of HgBr2 formed based on the limiting reagent.

5. Calculate the mass of HgBr2 formed.

Remember to double-check your calculations and ensure all units are consistent throughout the calculations.

I hope this helps you understand how to solve these problems! Let me know if you have any further questions.