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?

Question

asked Jan 23, 2021 234k views

1 Answer

Ulrich Dohou · Answer 1 · 2021-01-28T14:45:32+0000

Mass of CaBr₂ : 80 g

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 :

$\tt 0.4* 200=80~g$