Question

Рассчитайте, сколько должна весить накипь, выпавшая при выпаривании 100 л воды, если жесткость обусловлена только присутствием гидрокарбоната кальция и равна 5 мг-экв/л

Answer 1

M(Ca(HCO3)2=10*162/40=40,5 г

Answer 2

Calculation of Scale Weight

To calculate the weight of scale formed during the evaporation of 100 liters of water, we need to consider the hardness of the water, which is determined by the presence of calcium hydrocarbonate. The given hardness is 5 mg-eq/L.

Calculation: 1. Determine the molar mass of calcium hydrocarbonate (Ca(HCO3)2): - Calcium (Ca) has a molar mass of 40.08 g/mol. - Hydrogen (H) has a molar mass of 1.01 g/mol. - Carbon (C) has a molar mass of 12.01 g/mol. - Oxygen (O) has a molar mass of 16.00 g/mol. - The molar mass of calcium hydrocarbonate is calculated as follows: ``` (40.08 g/mol) + 2 * [(1.01 g/mol) + (12.01 g/mol) + (3 * (16.00 g/mol))] = 162.12 g/mol ```

2. Calculate the number of moles of calcium hydrocarbonate in 1 liter of water: - The hardness of the water is given as 5 mg-eq/L, which means 5 milliequivalents of calcium hydrocarbonate per liter of water. - Since 1 milliequivalent (meq) is equal to 0.001 moles, we can calculate the number of moles of calcium hydrocarbonate as follows: ``` 5 mg-eq/L * (0.001 mol/mg-eq) = 0.005 mol/L ```

3. Calculate the number of moles of calcium hydrocarbonate in 100 liters of water: - Multiply the number of moles per liter by the volume of water: ``` 0.005 mol/L * 100 L = 0.5 mol ```

4. Calculate the weight of calcium hydrocarbonate: - Multiply the number of moles by the molar mass of calcium hydrocarbonate: ``` 0.5 mol * 162.12 g/mol = 81.06 g ```

Therefore, the weight of scale formed during the evaporation of 100 liters of water, considering the given hardness of 5 mg-eq/L, is approximately 81.06 grams.