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

May 11, 2017

We need to assess the stoichiometric equation............and get a mass of approx. $125 \cdot g$ with respect to $\text{calcium bromide}$.

#### Explanation:

$C a \left(s\right) + B {r}_{2} \left(l\right) \rightarrow C a B {r}_{2} \left(s\right)$

$\text{Moles of metal} = \frac{25.0 \cdot g}{40.1 \cdot g \cdot m o {l}^{-} 1} = 0.623 \cdot m o l$.

And because of the molar equivalence, as expressed by the stoichiometric equation, $\text{moles of metal "-=" moles of salt}$.

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?