Asked by IRA
How many grams of Ca(OH)2 are needed to neutralize 5.00 g of 3-methylbutanoic acid?
a) 4.20 g
b) 2.10 g
c) 3.63 g
d) 1.81 g
a) 4.20 g
b) 2.10 g
c) 3.63 g
d) 1.81 g
Answers
Answered by
DrBob222
Let's call the acid HBu
2HBu + Ca(OH)2 ==> 2H2O + Ca(Bu)2
How many moles do you have in 5 g of the acid. That's
mols = grams/molar mass = 5.0/molar mass of the acid.
You see from the equation that 1 mol Ca(OH)2 required TWO mols of the acid. So mols Ca(OH)2 = 1/2 x mols acid
Then grams of the Ca(OH)2 = mols Ca(OH)2 x molar mass Ca(OH)2 = ?
2HBu + Ca(OH)2 ==> 2H2O + Ca(Bu)2
How many moles do you have in 5 g of the acid. That's
mols = grams/molar mass = 5.0/molar mass of the acid.
You see from the equation that 1 mol Ca(OH)2 required TWO mols of the acid. So mols Ca(OH)2 = 1/2 x mols acid
Then grams of the Ca(OH)2 = mols Ca(OH)2 x molar mass Ca(OH)2 = ?
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