Determine how many milliliters of 0.370 M HClO4 will be required to neutralize 58.99 g Ca(OH)2 according to the reaction:
2HC1O4+Ca(OH)2>CaC1O4)2+2H2O
How would I set this up?
To determine how many milliliters of 0.370 M HClO4 will be required to neutralize 58.99 g of Ca(OH)2, you need to follow these steps:
Step 1: Convert the mass of Ca(OH)2 to moles.
To do this, you need to know the molar mass of Ca(OH)2.
Ca(OH)2 has a molar mass of:
Ca = 40.08 g/mol
O = 16.00 g/mol (there are two oxygen atoms)
H = 1.01 g/mol (there are two hydrogen atoms)
So, the molar mass of Ca(OH)2 is:
40.08 g/mol + (2 * 1.01 g/mol) + (16.00 g/mol * 2) = 74.10 g/mol
To convert the mass of Ca(OH)2 to moles, divide the given mass by the molar mass:
58.99 g ÷ 74.10 g/mol = 0.795 mol
Step 2: Determine the stoichiometry of the reaction.
From the balanced equation, you can see that it takes 2 moles of HClO4 to neutralize 1 mole of Ca(OH)2.
Step 3: Use the stoichiometry to calculate the moles of HClO4 needed.
Since it takes 2 moles of HClO4 to neutralize 1 mole of Ca(OH)2, you can set up a ratio:
2 moles HClO4 / 1 mole Ca(OH)2 = x moles HClO4 / 0.795 moles Ca(OH)2
Cross-multiplying and solving for x, you'll get:
x = (2 moles HClO4 / 1 mole Ca(OH)2) * 0.795 moles Ca(OH)2
x = 1.59 moles HClO4
Step 4: Convert moles of HClO4 to milliliters.
To convert moles of HClO4 to milliliters, you need to use its molarity (0.370 M) and the equation: Molarity (M) = moles ÷ volume (in liters).
Rearrange the equation to solve for volume (in liters):
Volume (in liters) = moles / molarity
Volume (in liters) = 1.59 moles HClO4 / 0.370 M = 4.2973 liters
Finally, convert liters to milliliters by multiplying by 1000:
4.2973 liters * 1000 = 4297.3 milliliters
Therefore, you would need approximately 4297.3 milliliters of 0.370 M HClO4 to neutralize 58.99 g of Ca(OH)2.