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# How many grams of $Br$ are in $195g$ of $CaB{r_2}$ ?

Last updated date: 22nd Feb 2024
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Hint: In request to sort out the number of grams of bromine you get in that numerous grams of calcium bromide, $CaB{r_2}$ , you should locate the compound's percent creation.
To do that, utilization the way that one mole of calcium bromide contains
-One mole of calcium cations, $C{a^{2 + }}$
-Two moles of bromide anions, $2 \times B{r^ - }$

This will be a typical subject too! The mantra you should adhere to is grams to moles, to moles, to grams. How about we stroll through that. We will take the grams of $CaB{r_2}$ and convert it to moles. At that point from moles of $CaB{r_2}$ , we'll convert that to moles of $Br$ , and afterward convert THAT to grams of $Br$ .
You can in this manner use the molar mass of calcium bromide and the molar mass of bromine to decide the number of grams of bromine you get per $100{\text{ }}g$ of calcium bromide.
The two molar mass are
${\text{For CaB}}{{\text{r}}_2}:\;\:\;\:{M_M} = {\text{199}}{\text{.89 g mo}}{{\text{l}}^{ - 1}}$
${\text{For Br:}}\;\:\;\:\;\:\;\:\;\:\;\:{M_M} = {\text{79}}{\text{.904 g mo}}{{\text{l}}^{ - 1}}$
In this way, two moles of bromide anions for each one mole of calcium bromide will give you a percent creation of
$\dfrac{{2 \times 79.904\not{{{\text{g mo}}{{\text{l}}^{ - 1}}}}}}{{199.89\not{{{\text{g mo}}{{\text{l}}^{ - 1}}}}}} \times 100 = {\text{79}}{\text{.95% Br}}$
This implies that each $100{\text{ }}g$ of calcium bromide will contain $79.95{\text{ }}g$ of essential bromine as bromide cations.
All you need to do now is utilize this percent piece as a change factor to decide the number of grams of bromine you get in that $195 - g$ test of calcium bromide
$195\not{{{\text{g CaB}}{{\text{r}}_2}}} \cdot \mathop {\mathop {\dfrac{{{\text{79}}{\text{.95 g Br}}}}{{100\not{{{\text{g CaB}}{{\text{r}}_2}}}}}}\limits^ \downarrow }\limits^{{\text{79}}{\text{.95% Br}}} = \left[ {156gBr} \right]$
The appropriate response is adjusted to three sig figs.

Note: Mole percent is the rate that the moles of a specific segment are of the all-out moles that are in a combination. we utilize the molar mass of $CaB{r_2}$ to get to moles, and utilize the addendum $2$ to get to moles of $Br$ from moles of $CaB{r_2}$ , and afterward the molar mass of $Br$ to get to grams of $Br$ !