AgNO_{3} + NaBr → AgBr + NaNO_{3}

Calculating the mol of KBr in each solution

$36\overline{)\mathrm{mL}}\left(\frac{1\overline{)\mathrm{L}}}{1\mathrm{x}{10}^{3}\overline{)\mathrm{mL}}}\right)\left(\frac{1.30\mathrm{mol}}{1\overline{)\mathrm{L}}}\right)=0.0463\mathrm{mol}\mathrm{KBr}\phantom{\rule{0ex}{0ex}}61\overline{)\mathrm{mL}}\left(\frac{1\overline{)\mathrm{L}}}{1\mathrm{x}{10}^{3}\overline{)\mathrm{mL}}}\right)\left(\frac{0.610\mathrm{mol}}{1\overline{)\mathrm{L}}}\right)=0.03721\mathrm{mol}\mathrm{KBr}$

A 36.0 mL sample of 1.30 M KBr and a 61.0 mL sample of 0.610 M KBr are mixed. The solution is then heated to evaporate water until the total volume is 55.0 mL .

How many grams of silver nitrate are required to precipitate out silver bromide in the final solution?

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