Question

A reaction of 56.9 g of Na and 57.3g of Br2 yields 65.0g of NaBr. What is the percent yield?

2Na(s)+Br2(g) >> 2NaBr(s)

Answer 1

number of moles of Na = mass of Na / molar mass of Na
= 56.9/23
= 2.474 mol

number of moles of Br2 = mass of Br2 / molar mass of Br2
= 65.0/159.8
= 0.407 mol

2 mol of Na reacts with 1 mol of Br2
Br2 is present in less amount . SO Br2 is limiting reagent

from reaction,
moles of NaBr formed = 2*moles of Br2
= 2*0.407 mol
=0.814 mol

molar mass of NaBr = 102.9 g/mol

theoretical mass of NaBr = molar mass * number of moles
= 102.9 * 0.814
= 83.8 g

actual yield = 65.0 g

percent yield = actual *100 / theoretical
= 65.0*100/83.8
= 77.6 %
Answer: 77.6 %