1. Write the balanced chemical equation:

AgNO₃(aq) + NaBr(aq) → AgBr(s) + NaNO₃(aq)

2. Determine the limiting reactant:

* Calculate moles of AgNO₃:

(5.00 mL) * (0.100 mol/L) * (1 L/1000 mL) = 0.000500 mol AgNO₃

* Calculate moles of NaBr: You didn't provide a concentration for the sodium bromide solution, so I'll assume it's a 0.200 M solution for this example.

We need the volume of the NaBr solution to calculate moles. Let's assume we have 10.0 mL of the NaBr solution.

(10.0 mL) * (0.200 mol/L) * (1 L/1000 mL) = 0.00200 mol NaBr

* The limiting reactant is AgNO₃ because it has fewer moles than NaBr. This means that AgNO₃ will be completely consumed in the reaction, and the amount of AgBr formed will be determined by the amount of AgNO₃.

3. Calculate the moles of AgBr formed:

Since the balanced equation shows a 1:1 mole ratio between AgNO₃ and AgBr, the moles of AgBr formed will be equal to the moles of AgNO₃ used:

0.000500 mol AgNO₃ = 0.000500 mol AgBr

4. Calculate the mass of AgBr formed:

* Find the 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

* Calculate the mass:

(0.000500 mol AgBr) * (187.77 g/mol) = 0.0939 g AgBr

Therefore, approximately 0.0939 grams of AgBr will be formed.

Important Note: This calculation assumes the volume of the NaBr solution is 10.0 mL. If you have a different volume, you will need to recalculate the moles of NaBr and determine the limiting reactant accordingly.