Question

20 gram Ca and 64 gram Br reacts to form CaBr2. How many grams of CaBr2 form if 20% of Ca remains at the end of the reaction?

Answer 1

Mass of CaBr₂ : 80 g

Further explanation

Reaction

Ca+2Br⇒CaBr₂

mass of Ca = 20 g

mol of Ca (MW=40 g/mol):

`\tt (20)/(40)=0.5`

mass of Br = 64 g

mol Br(80 g/mol) :

`\tt (64)/(80)=0.8`

Ca remains at the end of the reaction⇒ Ca as an excess reactant

20% Ca remains(unreacted) :

`\tt 0.2* 20~g=4~g`

Ca reacted :

`\tt 20-4=16~g`

mol Ca reacted :

`\tt (16)/(40)=0.4`

mol CaBr₂ = mol Ca reacted = 0.4

mass CaBr₂ (MW=200 g/mol) produced :

`\tt 0.4* 200=80~g`

Or you can use mol ratio from equation :

mol CaBr₂ : mol Br (as limiting reactant) = 1 : 2, so mol CaBr₂ :

`\tt (1)/(2)* 0.8=0.4`

mass CaBr₂ (MW=200 g/mol) produced :

`\tt 0.4* 200=80~g`