How much calcium bromide could be prepared given a mass of #25.0*g# with respect to the metal, and stoichiometric bromine?

1 Answer
May 11, 2017

Answer:

We need to assess the stoichiometric equation............and get a mass of approx. #125*g# with respect to #"calcium bromide"#.

Explanation:

#Ca(s) +Br_2(l)rarr CaBr_2(s)#

#"Moles of metal"=(25.0*g)/(40.1*g*mol^-1)=0.623*mol#.

And because of the molar equivalence, as expressed by the stoichiometric equation, #"moles of metal "-=" moles of salt"#.

And thus we have a mass of #0.623*molxx199.89*g*mol=??g#

Is the reaction as written an example of a redox equation? Why or why not?