Hard water contains 80 mg / l of calcium bicarbonate and 20 mg / l of magnesium

| August 6, 2021 | education

Hard water contains 80 mg / l of calcium bicarbonate and 20 mg / l of magnesium bicarbonate. What mass of water is required to reduce 1 liter of such water

Given:
K (Ca (HCO3) 2) = 80 mg / l
K (Mg (HCO3) 2) = 20 mg / l
V (H2O) = 1 L

To find:
m (Na2CO3) -?

1) Ca (HCO3) 2 + Na2CO3 => 2NaHCO3 + CaCO3;
Mg (HCO3) 2 + Na2CO3 => 2NaHCO3 + MgCO3;
2) m (Ca (HCO3) 2) = K (Ca (HCO3) 2) * V (H2O) = 80 * 1 = 80 mg;
3) n (Ca (HCO3) 2) = m / M = 80/162 = 0.49 mmol;
4) n1 (Na2CO3) = n (Ca (HCO3) 2) = 0.49 mmol;
5) m1 (Na2CO3) = n1 * M = 0.49 * 106 = 51.94 mg;
6) m (Mg (HCO3) 2) = K (Mg (HCO3) 2) * V (H2O) = 20 * 1 = 20 mg;
7) n (Mg (HCO3) 2) = m / M = 20/146 = 0.14 mol;
8) n2 (Na2CO3) = n (Mg (HCO3) 2) = 0.14 mol;
9) m2 (Na2CO3) = n2 * M = 0.14 * 106 = 14.84 g;
10) m total. (Na2CO3) = m1 (Na2CO3) + m2 (Na2CO3) = 51.94 + 14.84 = 66.78 mg.

Answer: The mass of Na2CO3 is 66.78 mg.



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