Question

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

Answer 1

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?