A chemist reacts 45.0 mL of 5.6 M HCl with an excess of magnesium hydroxide Mg(OH)2. How many grams of magnesium chloride will be produced?
2HCl + Mg(OH)2 = MgCl2 + 2H2O
Since you have an excess of Mg (OH)2 - your limiting reagent is HCl.
In order to find out mass of MgCl2, you need to find out moles of MgCl2 first, which you will convert through a reaction from its reactant HCl (since it is a limiting reagents).
mole (HCl) = M x L = 45 mL x 5.6 M = 0.045 L x 5.6 M = 0.252 moles
HCl and MgCl2 are reacting in 2 : 1 ration - based on the balanced reaction above.
2 moles (HCl by reaction) - 0.252 moles (HCl practically)
1 moles (MgCl2 by reaction) - x moles (MgCl2 practically)
x (MgCl2) = 1 mole (MgCl2) x 0.252 moles (HCl) / 2 moles (HCl) = 0.126 moles of MgCl2
Molar Mass of MgCl2 = 24+71 = 95 g/mol
Now, mass (MgCl2) = moles x Molar mass = 0.126 moles x 95 g/mol = 11.97 grams
Final answer : m (MgCl2) = 11.97 grams
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