Question

A reaction of 42.7g of Na and 56.3g of Br2 yields 68.8g of NaBr. What is the percent yield?

2Na(s) + Br₂(g) ---> 2NaBr(s)

Answer 1

Atoimic weight of Sodium (Na)= 23 and Molecular weight of Br2= 160

moles = mass/atomic weight in case of Na, Moles of Na=42.7/ 23=1.8565 moles

Moles of Br2= Mass/Molecular weight= 56.3/160=0.352 moles

Theoreteical moles ratio of Na:Br2= 2:1

Actial moles ratio taken = 1.8565: 0.352 = 5.3 :1

Sodium is the excess reactant ( since ii is 5.3 :1 against 2:1)

Br2 is the limting reactantt and determines the extetnt of foramtion of NaBr

1 mole of Br2 gives 2 moles of NaBr ( theoretical yiled)

0.352 moles gives 2*0.352 moles of NaBr =0.704 moles of NaBr. This is the theoretical yiled that can be expected.

Molecular weight of NaBr= 23+80= 103, mass of the theoretical yield =moles* Molecular weight= 0.704*103=72.5 gm

But actual yiled= 68.8 gms

Percent yield = 100* actual yiled/ Theoretical yiled= 100*68.8/72.5=94.89%