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

Question

Serenity Allen · Accepted Answer

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

#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# Why or why not does the reaction as written represent an example of a redox equation?

Cooper Anderson · Answer

undefined To calculate the amount of calcium bromide that could be prepared, we first need to determine the molar mass of calcium bromide (CaBr2), which is 199.89 g/mol. Next, we calculate the moles of calcium bromide using the given mass of 25.0 g and the molar mass.

$$ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{25.0 g}{199.89 g/mol} = 0.125 \text{ mol} $$

Since calcium bromide has a 1:2 ratio with respect to calcium to bromine, we need twice as many moles of bromine as moles of calcium bromide.

$$ \text{moles of bromine} = 2 \times \text{moles of calcium bromide} = 2 \times 0.125 \text{ mol} = 0.250 \text{ mol} $$

Now, we use the molar mass of bromine (Br), which is 79.90 g/mol, to calculate the mass of bromine needed.

$$ \text{mass of bromine} = \text{moles of bromine} \times \text{molar mass of bromine} = 0.250 \text{ mol} \times 79.90 \text{ g/mol} = 19.975 \text{ g} $$

Therefore, if 25.0 g of calcium is reacted with stoichiometric amounts of bromine, 19.975 g of bromine would be consumed, resulting in the formation of 25.0 g of calcium bromide.