Question

In the following reaction, how many grams of NaBr will produce 244 grams of NaNO3? Pb(NO3)2(aq) 2 NaBr(aq) PbBr2(s) 2 NaNO3(aq) The molar mass of NaBr is 102.9 grams and that of NaNO3 is 85 grams

Answer 1

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