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
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 = ?
To determine the number of grams of Ca(OH)2 needed to neutralize 5.00 g of 3-methylbutanoic acid, we need to use the balanced chemical equation for the reaction between Ca(OH)2 and 3-methylbutanoic acid.
The balanced chemical equation is:
2 Ca(OH)2 + 2 C5H10O2 -> 2 Ca(C5H10O2) + 2 H2O
From the balanced chemical equation, we can see that the molar ratio between Ca(OH)2 and 3-methylbutanoic acid is 2:1. This means that for every 2 moles of Ca(OH)2, we need 1 mole of 3-methylbutanoic acid.
Now, let's calculate the molar mass of 3-methylbutanoic acid (C5H10O2):
M(C) = 12.01 g/mol
M(H) = 1.01 g/mol
M(O) = 16.00 g/mol
Molar mass of C5H10O2 = (5 * M(C)) + (10 * M(H)) + (2 * M(O))
= (5 * 12.01) + (10 * 1.01) + (2 * 16.00)
= 72.05 + 10.10 + 32.00
= 114.15 g/mol
Now, let's calculate the molar mass of Ca(OH)2:
M(Ca) = 40.08 g/mol
M(O) = 16.00 g/mol
M(H) = 1.01 g/mol
Molar mass of Ca(OH)2 = (1 * M(Ca)) + (2 * M(O)) + (2 * M(H))
= 40.08 + 32.00 + 2.02
= 74.10 g/mol
Now, let's calculate the number of moles of 3-methylbutanoic acid:
moles of 3-methylbutanoic acid = mass / molar mass
= 5.00 g / 114.15 g/mol
≈ 0.0438 mol
According to the balanced chemical equation, we need 2 moles of Ca(OH)2 to neutralize 1 mole of 3-methylbutanoic acid. Therefore, we need:
moles of Ca(OH)2 = (2 moles / 1 mole) * moles of 3-methylbutanoic acid
≈ 0.0438 mol * (2/1)
≈ 0.0876 mol
Finally, let's calculate the mass of Ca(OH)2 needed:
mass of Ca(OH)2 = moles of Ca(OH)2 * molar mass
= 0.0876 mol * 74.10 g/mol
≈ 6.49 g
The answer is not one of the given options.
Therefore, none of the options (a), (b), (c), or (d) are correct.
To find out how many grams of Ca(OH)2 are needed to neutralize 5.00 g of 3-methylbutanoic acid, we need to use the concept of stoichiometry.
First, we need to determine the balanced chemical equation for the neutralization reaction between 3-methylbutanoic acid and Ca(OH)2. The neutralization reaction can be written as:
2 CH3(CH2)2COOH + Ca(OH)2 -> Ca(CH3(CH2)2COO)2 + 2 H2O
From the balanced equation, we can see that 2 moles of 3-methylbutanoic acid react with 1 mole of Ca(OH)2 to produce 1 mole of Ca(CH3(CH2)2COO)2.
To calculate the moles of 3-methylbutanoic acid, we divide the given mass of 5.00 g by the molar mass of 3-methylbutanoic acid. The molar mass of 3-methylbutanoic acid (C6H12O2) can be calculated as:
(6 * 12.01 g/mol) + (12 * 1.01 g/mol) + (2 * 16.00 g/mol) = 116.15 g/mol
Number of moles of 3-methylbutanoic acid = 5.00 g / 116.15 g/mol
Now, we can use the stoichiometry from the balanced equation to calculate the moles of Ca(OH)2 needed. As per the balanced equation, 2 moles of 3-methylbutanoic acid react with 1 mole of Ca(OH)2. Therefore, the moles of Ca(OH)2 can be calculated as:
Moles of Ca(OH)2 = ( moles of 3-methylbutanoic acid ) / 2
Finally, to calculate the grams of Ca(OH)2 needed, we multiply the moles of Ca(OH)2 by its molar mass. The molar mass of Ca(OH)2 (calcium hydroxide) can be calculated as:
(1 * 40.08 g/mol) + (2 * 1.01 g/mol) + (2 * 16.00 g/mol) = 74.10 g/mol
Grams of Ca(OH)2 needed = ( moles of Ca(OH)2 ) * 74.10 g/mol
Now, let's calculate the grams of Ca(OH)2 needed:
Moles of 3-methylbutanoic acid = 5.00 g / 116.15 g/mol = 0.043 mol
Moles of Ca(OH)2 = 0.043 mol / 2 = 0.022 mol
Grams of Ca(OH)2 needed = 0.022 mol * 74.10 g/mol = 1.63 g
Therefore, the correct option is not given in the answer choices. The correct answer would be approximately 1.63 g, which is not listed.