1. what reaction take place when water is passed through anion exchange resin ?
2. how much na2co3 or salked lime should be added to completely 300cm^3 of 2percent cacl2 solution ?
1. When water is passed through an anion exchange resin, an ion exchange reaction occurs. Anion exchange resins contain negatively charged functional groups that attract and exchange negatively charged ions in the water. During this process, anions from the water, such as chloride (Cl-), nitrate (NO3-), or sulfate (SO42-), are replaced by other anions on the resin, such as hydroxide (OH-) or bicarbonate (HCO3-). This results in the removal of unwanted anions from the water and the release of the exchanged anions into the water.
2. To find out how much Na2CO3 or slaked lime should be added to completely react with 300 cm^3 of a 2% CaCl2 solution, we need to calculate the amount of moles of CaCl2 present in the solution and then determine the stoichiometric ratio between the CaCl2 and Na2CO3 or slaked lime.
To calculate the moles of CaCl2, we need to know its molar mass. Calcium has a molar mass of approximately 40.08 g/mol, while chlorine has a molar mass of approximately 35.45 g/mol. Therefore, the molar mass of CaCl2 is approximately 40.08 + (2 * 35.45) = 110.98 g/mol.
Next, we need to calculate the amount of CaCl2 in the solution using its concentration (2%) and the volume (300 cm^3). The concentration is given as a percentage, so we need to convert it to g/cm^3 by dividing by 100. Given that the density of water is approximately 1 g/cm^3, the concentration of CaCl2 is 2/100 * 1 g/cm^3 = 0.02 g/cm^3.
The mass of CaCl2 in the solution can be calculated by multiplying the concentration by the volume:
mass = concentration * volume = 0.02 g/cm^3 * 300 cm^3 = 6 g.
To calculate the moles of CaCl2, we divide the mass by the molar mass:
moles = mass / molar mass = 6 g / 110.98 g/mol ≈ 0.054 mol.
Now, we need to determine the stoichiometric ratios between CaCl2 and Na2CO3 or slaked lime. The balanced chemical equations for the reactions are:
CaCl2 + Na2CO3 -> 2 NaCl + CaCO3
CaCl2 + Ca(OH)2 -> 2 CaClOH + CO2
From the balanced equations, we can see that 1 mole of CaCl2 reacts with either 1 mole of Na2CO3 or 1 mole of Ca(OH)2.
Therefore, to completely react with 0.054 mol of CaCl2, we would need an equal amount of Na2CO3 or slaked lime.