**Step 1. Start with the balanced equation.**

If there is such a process, the balanced equation will probably be

#M_text(r):color(white)(m) 187.77color(white)(mmmml)102.89#

#color(white)(mmm)n"AgBr" + ? → n"NaBr" + ?#

where #n# is the same number on each side of the equation.

**Step 2. Calculate the moles of #"AgBr"#**.

#"Moles of AgBri" = 42.7 color(red)(cancel(color(black)("g AgBr"))) × ("1 mol AgBr")/(187.7 color(red)(cancel(color(black)("g AgBr")))) = "0.2275 mol AgBr"#

**Step 3. Calculate the moles of #"NaBr"#**

#"Moles of NaBr" = 0.2275 color(red)(cancel(color(black)("mol AgBr"))) × (stackrelcolor(blue)(1)(color(red)(cancel(color(black)(n)))) "mol NaBr")/(color(red)(cancel(color(black)(n))) color(red)(cancel(color(black)("mol AgBr")))) = "0.2275 mol NaBr"#

**4. Calculate the mass of #"NaBr"#**.

#"Mass of H"_2"O" = 0.2275 color(red)(cancel(color(black)("mol NaBr"))) × "102.89 g NaBr"/(1 color(red)(cancel(color(black)("mol NaBr")))) = "23.4 g NaBr"#