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

295.38 grams 


Hope this helped :)

Answer 2

Answer:

295.3744 grams of NaBr will produce 244 grams of $NaNO_3$.

Explanation:

$Pb(NO_3)_2(aq)+2 NaBr(aq)\rightarrow PbBr_2(s)+2 NaNO_3(aq)$

Moles of sodium nitarte= $\frac{244 g}{85 g/mol}=2.8705 mol$

According to reaction, 2 moles of sodium nitrate is obtained from 2 moles of sodium bromide.

Then 2.8705 mol of sodium nitrate will be obtained from :

$\frac{2}{2}\times 2.8705mol=2.8705 mol$ of sodium nitrate

Mass of 2.8705 moles of sodium nitrate:

$2.8705 mol\times 102.9 g/mol=295.3744 g$

295.3744 grams of NaBr will produce 244 grams of $NaNO_3$.