Question

A 0.9157-g mixture of CaBr2 and NaBr is dissolved in water, and AgNO3 is added to the solution to form AgBr precipitate. If the mass of the precipitate is 1.6900 g, what is the percent by mass of NaBr in the original mixture?

Answer 1

Answer: NaBr is 61.79% by mass.

Explanation: Reaction of silver nitrate with calcium bromide and sodium bromide are written as:

$CaBr_2(aq)+2AgNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+2AgBr(s)$

$NaBr(aq)+AgNO_3(aq)\rightarrow NaNO_3(aq)+AgBr(s)$

Let's say the mass of NaBr in the original solution is w grams. Then mass of $CaBr_2$ will be 0.9157 - w grams.

Molar mass of sodium bromide is 102.89 gram per mol, molar mass of calcium bromide is 199.89 gram per mol and molar mass of AgBr is 187.77 gram per mol.

moles of NaBr = $\frac{w}{102.89}$

moles of $CaBr_2$ = $\frac{0.9157-w}{199.89}$

moles of AgBr = $\frac{1.6900}{187.77}$  = 0.0090

From the balanced equations, 1 mol of calcium bromide gives 2 moles of silver bromide and 1 mol of sodium bromide gives 1 mol of silver bromide.

So, total moles of silver bromide = 2(moles of calcium bromide) + moles of sodium bromide

Let's plug in the values in it:

$0.0090=2(\frac{0.9157-w}{199.89})+\frac{w}{102.89}$

On solving this:

w = 0.5658

So, mass of NaBr in the original mixture is 0.5658 grams.

Mass percentage of NaBr in the original mixture = $(\frac{0.5658}{0.9157})*100$

= 61.79%