Answer:
276g of Br are present
Explanation:
To solve this question we need to find the moles of CaBr2 using its molar mass -CaBr2: 199.89g/mol-. As 1 mole of CaBr2 contains 2 moles of Br we can find the moles of Br and its mass:
<em>Moles CaBr2:</em>
345g * (1mol/199.89g) = 1.7259 moles CaBr2
<em>Moles Br:</em>
1.7259 moles CaBr2 * (2mol Br / 1mol CaBr2) = 3.4519 moles Br
<em>Mass Br -Molar mass: 79.904g/mol-</em>
3.4519 moles Br * (79.904g/mol) = 276g of Br are present
Please note that since magnesium is in group 2 it has a valency of 2+ and so the formula for magnesium hydroxide is
M
g
(
O
H
)
2
[Mistake in question]
This is a typical neutralization reaction of an acid with a base to form a salt and water. The reaction is exothermic, gives off heat,
Δ
H
<
0
, and may be balanced by adding balancing numbers in front, ie adding molecules, in order to ensure that the total number of atoms of each element is the same on the left and right hand sides of the equation.
Doing so we obtain :
2
H
3
P
O
4
+
3
M
g
(
O
H
)
2
→
M
g
3
(
P
O
4
)
2
+
6
H
2
O
