Answer : The mass of \(NaBr\) is, 295.323 grams
Solution :
First we have to calculate the moles of \(NaNO_3\).
\(\text{Moles of }NaNO_3=\frac{\text{Mass of }NaNO_3}{\text{Molar mass of }NaNO_3}=\frac{244g}{85g/mole}=2.87moles\)
Now we have to calculate the moles of NaBr.
The balanced chemical reaction is,
\(Pb(NO_3)_2(aq)+2NaBr(aq)\rightarrow PbBr_2(s)+2NaNO_3(aq)\)
From the balanced reaction we conclude that
As, 2 moles of \(NaNO_3\) react with 2 moles of \(NaBr\)
So, 2.87 moles of \(NaNO_3\) react with 2.87 moles of \(NaBr\)
Now we have to calculate the mass of NaBr.
\(\text{Mass of }NaBr=\text{Moles of }NaBr\times \text{Molar mass of }NaBr\)
\(\text{Mass of }NaBr=(2.87mole)\times (102.9g/mole)=295.323g\)
Therefore, the mass mass of \(NaBr\) is, 295.323 grams
