Mass of CaBr₂ : 80 g
Reaction
Ca+2Br⇒CaBr₂
mass of Ca = 20 g
mol of Ca (MW=40 g/mol):
[tex]\tt \dfrac{20}{40}=0.5[/tex]
mass of Br = 64 g
mol Br(80 g/mol) :
[tex]\tt \dfrac{64}{80}=0.8[/tex]
Ca remains at the end of the reaction⇒ Ca as an excess reactant
20% Ca remains(unreacted) :
[tex]\tt 0.2\times 20~g=4~g[/tex]
Ca reacted :
[tex]\tt 20-4=16~g[/tex]
mol Ca reacted :
[tex]\tt \dfrac{16}{40}=0.4[/tex]
mol CaBr₂ = mol Ca reacted = 0.4
mass CaBr₂ (MW=200 g/mol) produced :
[tex]\tt 0.4\times 200=80~g[/tex]
Or you can use mol ratio from equation :
mol CaBr₂ : mol Br (as limiting reactant) = 1 : 2, so mol CaBr₂ :
[tex]\tt \dfrac{1}{2}\times 0.8=0.4[/tex]
mass CaBr₂ (MW=200 g/mol) produced :
[tex]\tt 0.4\times 200=80~g[/tex]
