Question

A reaction of 35.1 g of Na and 33.7 g of Br2 yields 34.8 g of NaBr. What is the percent yield? 2Na(s)+Br2(g)-->2NaBr(s)

Answer 1

According to reaction 2 mole of Na react with 1 mole Br₂

molar mass of Na = 22.9897gm/mol then 2 mole of Na = 45.9794 gm

molar mass of Br₂ = 159.808gm/mole

that mean for complete reaction 45.9794 gm of Na require react with 159.808 gm Br₂

then 33.7 gm Br₂ require 33.745.9794/159.808 = 9.696 gm of Na

we have taken 35.1 gm Na thus Na is excess reactant some volume remain unreacted and Br₂ is limiting reactant react completly

molar mass of NaBr = 102.894 gm/mole there is 2 mole of NaBr formed so, 2NaBr = 205.788 gm

that mean 158.808 gm Br₂ produce 205.788 gm NaBr then 33.7 gm Br₂ produce 33.7205.788/158.808 = 43.669gm

43.669 gm NaBr = theoratical yield = 100% yield then 34.8 gm = 34.8100/43.669 = 79.69%

Percent yield = 79.69%