Answer: I hope this helps 0w0
Step-by-step explanation:
Explanation:
In order to figure out how many grams of bromine you get in that many grams of calcium bromide,
CaBr
2
, you must find the compound's percent composition.
To do that, use the fact that one mole of calcium bromide contains
one mole of calcium cations,
Ca
2
+
two moles of bromide anions,
2
×
Br
−
You can thus us the molar mass of calcium bromide and the molar mass of bromine to determine how many grams of bromine you get per
100 g
of calcium bromide.
The two molar mass are
For CaBr
2
:
M
M
=
199.89 g mol
−
1
For Br:
M
M
=
79.904 g mol
−
1
So, two moles of bromide anions for every one mole of calcium bromide will give you a percent composition of
2
×
79.904
g mol
−
1
199.89
g mol
−
1
×
100
=
79.95% Br
This means that every
100 g
of calcium bromide will contain
79.95 g
of elemental bromine in the form of bromide cations.
All you have to do now is use this percent composition as a conversion factor to determine how many grams of bromine you get in that
195-g
sample of calcium bromide
195
g CaBr
2
⋅
79.95% Br
79.95 g Br
100
g CaBr
2
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
156 g Br
a
a
∣
∣
−−−−−−−−−−−−