Question 3: When a sample of solid silver bromide decomposes over time it will form solid silver and bromine gas. This decomposition does reach completion and all reactant transitions to products.

Question

A. Write the full reaction with the correct physical states and balance the equation with all required work/steps or explanation, if needed. make a table for the atoms in the products and reactants in the current equation.

B. How many grams of bromine gas will be produced when 25.54 grams of silver bromide solid decomposes fully? Must show the correct stoichiometric set up for calculation.

Answer 1

A. The full reaction is:

AgBr(s) -> Ag(s) + Br2(g)

Balancing the equation:

2AgBr(s) -> 2Ag(s) + Br2(g)

Table for atoms:

| | Reactants | Products |
|:---------:|:---------:|:--------:|
| Silver (Ag) | 2 | 2 |
| Bromine (Br) | 2 | 2 |

B. To determine how many grams of bromine gas will be produced when 25.54 grams of silver bromide decomposes fully, we first need to calculate the molar mass of AgBr and Br2.

Molar mass of AgBr:
Ag: 107.87 g/mol
Br: 79.90 g/mol
Molar mass of AgBr = 107.87 + 79.90 = 187.77 g/mol

Using the molar mass of AgBr, we can calculate the moles of AgBr in 25.54 grams:
25.54 g / 187.77 g/mol = 0.136 mol AgBr

From the balanced equation, we see that 2 moles of AgBr produce 1 mole of Br2. Therefore, using stoichiometry:

0.136 mol AgBr x (1 mol Br2 / 2 mol AgBr) = 0.068 mol Br2

Finally, we convert moles of Br2 to grams by using the molar mass of Br2:
79.90 g/mol x 0.068 mol = 5.44 grams of bromine gas will be produced.