Calculating Mercury(II) Bromide Production: Mass of Reactants & Products

[suspenc3]

Mercury and bromine will react with each other to produce mercury(II) bromide.
a.)What mass is produced from the reaction of 10.0g Hg and 10.0g Br2?What mass of which reagent is left unreacted?

b.)what mass of HgBr2 is produced from the reaction of 5.00mL of mercury?(d=13.5g/mL) and 5.00mL of bromine(d=3.12g/mL)?

If someone could give me a clue where to start? My High school had a terrible chemistry program, I've never seen any of this before.

Please, any help would be appreciated.

 

[siddharth]

You can start by writing the reaction which takes place. That will tell you, how many moles of the product is formed per mole of the reactant consumed.
(Are you familiar with these terms?)

You can find more in the links below
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch3/equations.html
http://web.jjay.cuny.edu/~acarpi/NSC/6-react.htm

 

[Blueshift5]

a.) To answer this question, we need to first write out the balanced chemical equation for the reaction:

Hg + Br2 -> HgBr2

Next, we need to calculate the molar mass of each reactant and product. The molar mass of Hg is 200.59 g/mol and the molar mass of Br2 is 159.81 g/mol. The molar mass of HgBr2 is 298.61 g/mol.

Using the given masses of 10.0g Hg and 10.0g Br2, we can calculate the moles of each reactant by dividing the mass by its molar mass. This gives us 0.0499 moles of Hg and 0.0625 moles of Br2.

Next, we need to determine which reactant is the limiting reagent. This is the reactant that is completely used up in the reaction and determines the amount of product that can be formed. To do this, we compare the moles of each reactant to the stoichiometric ratio in the balanced equation. In this case, the ratio is 1:1 for Hg and Br2. This means that for every 1 mole of Hg, 1 mole of Br2 is needed. Since we have less moles of Hg, it is the limiting reagent.

To calculate the mass of HgBr2 produced, we use the limiting reagent (Hg) and the stoichiometric ratio. This gives us 0.0499 moles of HgBr2. To convert this to grams, we multiply by the molar mass of HgBr2, giving us 14.9g of HgBr2 produced.

To determine the mass of unreacted reagent, we subtract the moles of Hg used (0.0499 moles) from the initial moles of Hg (0.0499 moles). This gives us 0 moles of Hg left unreacted. Similarly, for Br2, we have 0.0625 moles used and 0.0625 moles left unreacted. To convert this to grams, we multiply by the molar mass of Br2, giving us 19.5g of Br2 left unreacted.

b.) To answer this question, we first need to calculate the volume of each reagent in liters.