Sure, let's work through this problem step-by-step.
1. Identify the molar masses of the compounds involved:
- Molar mass of AgBr (Silver Bromide) is given as 187.77 g/mol.
- Molar mass of NaBr (Sodium Bromide) is given as 102.89 g/mol.
2. Calculate the moles of NaBr produced:
- We start with 75.0 grams of NaBr.
- To find the number of moles of NaBr, we use the formula:
[tex]\[
\text{moles of NaBr} = \frac{\text{mass of NaBr}}{\text{molar mass of NaBr}} = \frac{75.0 \text{ g}}{102.89 \text{ g/mol}} \approx 0.7289 \text{ moles}
\][/tex]
3. Determine the moles of AgBr required:
- From the balanced chemical equation, the stoichiometry between AgBr and NaBr is 1:1. Therefore, the number of moles of AgBr required is the same as the number of moles of NaBr produced.
- Hence, the moles of AgBr required is also approximately 0.7289 moles.
4. Calculate the mass of AgBr required:
- To find the mass of AgBr, we use the formula:
[tex]\[
\text{mass of AgBr} = \text{moles of AgBr} \times \text{molar mass of AgBr} = 0.7289 \text{ moles} \times 187.77 \text{ g/mol} \approx 136.87 \text{ g}
\][/tex]
Therefore, approximately 136.87 grams of AgBr would be required to produce 75.0 grams of NaBr.
