# What is the percent yield for the reaction between 45.9 g of NaBr and excess chlorine gas to produce 12.8 g of NaCl and an unknown quantity of bromine gas?

Aug 7, 2017

Approx. 50%...........

#### Explanation:

We interrogate the given reaction.....

$N a B r \left(a q\right) + \frac{1}{2} C {l}_{2} \left(a q\right) \rightarrow N a C l \left(a q\right) + \frac{1}{2} B {r}_{2} \left(a q\right)$

Bromide ion is oxidized, and chlorine is reduced........

$\text{Moles of NaBr} = \frac{45.9 \cdot g}{102.89 \cdot g \cdot m o {l}^{-} 1} = 0.446 \cdot m o l$

$\text{Moles of NaCl} = \frac{12.8 \cdot g}{58.44 \cdot g \cdot m o {l}^{-} 1} = 0.219 \cdot m o l$

And the yield is simply the quotient of these molar quantities.....

"% Yield"="Moles of NaCl"/"Moles of NaBr"xx100%

=(0.219*mol)/(0.446*mol)xx100%=??%