Question

A reaction of 50.1 g of Na and 31.7 g of Br2 yields 33.4 g of NaBr. What is the percent yield? 2Na +Br2______2NaBr

Answer 1

Well, we write a stoichiometric equation to inform our accounting...

Explanation:

`Na(s) + 1/2Br_2(l) rarr NaBr(s)`

`"Moles of sodium"=(50.1*g)/(22.99*g*mol^-1)=2.18*mol`.

`"Moles of bromine"=(31.7*g)/(159.8*g*mol^-1)=0.198*mol`.

Sodium is present in stoichiometric excess...and we use the equivalent quantity of bromine to determine the yield .. AT MOST we can make `2xx0.198*molxx102.89*g*mol^-1=40.8*g` of salt....

And so `"% yield"=(33.4*g)/(40.8*g)xx100%=81.9%`