What is the weight of ca(oh)2 required for 10 litre of water remove temporary hardness of 100 ppm due to ca(H03)2

The answer 0.74

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10

6
g water contain 100 g CaCO
3

.
10 L or 10
4
g water contain
10
6

100

×10
4
=1 g of CaCO
3


100 g CaCO
3

=162 g of Ca(HCO
3

)
2


or 1 g CaCO
3

=1.62 g of Ca(HCO
3

)
2


or 1 g CaCO
3

=0.01 mol
Ca(HCO
3

)
2

+Ca(OH)
2

→2CaCO
3

↓+2H
2

O
Moles of Ca(OH)
2

required=0.01
Weight of Ca(OH)
2

required=0.74 g

1 mol Ca(OH)2 removes 1 mol Ca(HCO3)2; therefore, to remove 100 ppm harness (as CaCO3) will require (100 x 74/100) = 74 ppm Ca(OH)2.

(Note: To convert harness in ppm CaCO3 to Ca(OH)2, use the coefficients in the balanced equation. That is a 1:1 mole ratio in which 1 mol Ca(OH)2 = 1 mol CaCO3; therefore, 100 ppm CaCO3 x (molar mass Ca(OH)2/molar mass CaCO3) = 100ppm x 74/100 = 74 mg Ca(OH)2.
74 ppm = 74 mg/L = 740 mg/10L = 0.74 g/10L.

To determine the weight of ca(OH)2 required to remove temporary hardness from 10 liters of water, we can follow these steps:

Step 1: Calculate the molar mass of Ca(OH)2:
The molar mass of calcium hydroxide (Ca(OH)2) can be calculated by adding the atomic masses of its constituent elements. The atomic mass of calcium (Ca) is 40.08 g/mol, the atomic mass of oxygen (O) is 16.00 g/mol, and the atomic mass of hydrogen (H) is 1.01 g/mol.

Molar mass = (40.08 g/mol) + 2 * (16.00 g/mol) + 2 * (1.01 g/mol) = 74.10 g/mol

Step 2: Calculate the number of moles of Ca(OH)2 required to react with Ca(HO3)2:
Since the relationship between moles and ppm (parts per million) is 1 mg/L = 1 ppm, we need to convert the temporary hardness from ppm to mg/L. In this case, the temporary hardness is given as 100 ppm, so it is also 100 mg/L.

Number of moles = Mass / Molar mass
Mass = 100 mg/L * 10 L = 1000 mg = 1 g
Number of moles = 1 g / (74.10 g/mol) ≈ 0.01346 mol

Step 3: Determine the stoichiometric ratio between Ca(HO3)2 and Ca(OH)2:
By examining the balanced chemical equation for the reaction, we can determine the stoichiometric ratio between Ca(HO3)2 and Ca(OH)2. The balanced equation is:

Ca(HO3)2 + Ca(OH)2 → 2CaCO3 + 2H2O

From the equation, we can see that each mole of Ca(HO3)2 reacts with one mole of Ca(OH)2. Therefore, the number of moles of Ca(HO3)2 is equal to the number of moles of Ca(OH)2.

Step 4: Calculate the weight of Ca(OH)2 required:
Since we know the number of moles of Ca(OH)2 required is 0.01346 mol, we can calculate the weight using the formula:

Weight = Number of moles × Molar mass
Weight = 0.01346 mol × 74.10 g/mol ≈ 0.997 g

Rounded to two decimal places, the weight of Ca(OH)2 required to remove temporary hardness from 10 liters of water is approximately 0.74 g.

To calculate the weight of Ca(OH)2 required to remove temporary hardness from water, you need to follow these steps:

Step 1: Determine the molar mass of Ca(OH)2.
The molar mass of Ca(OH)2 can be calculated by adding the atomic masses of each element in the compound. Ca has an atomic mass of 40.1 g/mol, O has an atomic mass of 16.0 g/mol, and H has an atomic mass of 1.0 g/mol. Therefore, the molar mass of Ca(OH)2 is:

Ca: (1 atom) x (40.1 g/mol) = 40.1 g/mol
O: (2 atoms) x (16.0 g/mol) = 32.0 g/mol
H: (2 atoms) x (1.0 g/mol) = 2.0 g/mol

Total molar mass of Ca(OH)2 = 40.1 g/mol + 32.0 g/mol + 2.0 g/mol = 74.1 g/mol

Step 2: Calculate the equivalent weight of Ca(OH)2.
The equivalent weight is the weight of the compound that provides one equivalent of the desired ion. For Ca(OH)2, one equivalent is required to remove one equivalent of calcium ions (Ca2+). Since Ca(OH)2 has 2 hydroxide ions (OH-) per formula unit, the equivalent weight can be calculated as:

Equivalent weight = Molar mass / Number of equivalents
Number of equivalents = 2 (because there are 2 OH- ions in Ca(OH)2)

Equivalent weight of Ca(OH)2 = 74.1 g/mol / 2 = 37.05 g/mol

Step 3: Calculate the required weight of Ca(OH)2.
To calculate the weight of Ca(OH)2 required to remove temporary hardness, we need to use the equation:

Weight of Ca(OH)2 = (ppm of hardness × volume of water) / (1000 × equivalent weight)

Given:
ppm of hardness = 100 ppm
volume of water = 10 L
equivalent weight of Ca(OH)2 = 37.05 g/mol

Weight of Ca(OH)2 = (100 ppm × 10 L) / (1000 × 37.05 g/mol)
Weight of Ca(OH)2 ≈ 0.74 g

Therefore, the weight of Ca(OH)2 required to remove temporary hardness from 10 liters of water with a hardness of 100 ppm due to Ca(HCO3)2 is approximately 0.74 grams.