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You need to first write the balanced reaction between magnesium hydroxide and hydrobromic acid, such that:
`Mg(OH)_2 + 2HBr -> MgBr_2 + 2H_2O`
You need to solve the problem for the amount of magnesium bromide formed from the given 40g of hydrobromic acid, such that:
grams of `MgBr_2` = 40g `HBr` * (1mole `HBr` )/(atomic mass of `HBr` )*(1mole of `MgBr_2` )/(2 moles of `HBr` )*(atomic mass of `MgBr_2` )/(1 mole `MgBr_2` )
You need to use the periodic table to find out the needed atomic masses such that:
atomic mass of` HBr = 1 g H + 79 g Br = 80` grams
atomic mass of `MgBr_2 = 24 g Mg + 2*79 g Br = 24 g + 158 g = 182 ` grams
Replacing the values in the formula above, yields:
grams of `MgBr_2 = (40*1*1*182)/(80*2*1) = 182/4 = 45.5` grams
Hence, evaluating the amount of magnesium bromide, under the given conditions, yields `45.5` grams of `MgBr_2` .
First, we need to write the basic chemical equation for the reaction of hydrobromic acid with magnesium hydroxide.
`2HBr + Mg(OH)_2 -> 2H_2O + MgBr_2`
Next, we need to do some stoichiometry in order to determine the result from reacting 40.0ml of hydrobromic acid. We can take the density of the acid to be equal to approximately 1, thus we can change 40.0ml to 40.0g.
Now we can use stoichiometric coefficents from the equation and molar masses of the compounds to determine how much magnesium bromide will be produced.
`40.0g HBr * ((1mol HBr)/(81g HBr)) * ((1mol MgBr_2)/(2mol HBr)) * ((184g MgBr_2)/(1mol MgBr_2)) = 45.43g MgBr_2`
Thus, 45.43g of magnesium bromide will be produced.
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