Chemistry · Chemical equations and stoichiometry
of sample of hard water gave good lather with of standard soap solution (1 mL so
\( 50 \mathrm{mL} \) of sample of hard water gave good lather with \( 6 \mathrm{mL} \) of standard soap solution (1 mL soap solutions = 1 mg \( \mathrm{CaCO}_{3} \). If the hardness is only due to \( M g\left(H C O_{3}\right)_{2}, \) the weight of milk of lime required to remove the hardness completely from \( 100 \mathrm{kg} \) of that sample of water is:
- A. \( 17.8 \mathrm{g} \)
- B. \( 8.9 \mathrm{g} \)
- C. \( 178 g \)
- D. 89 \( g \)
Step-by-step solution
The hardness of water is 6 mg CaCO3 equivalent per 50 mL, so 120 mg/L. This corresponds to 0.0012 mol/L of Mg(HCO3)2. Reaction requires equimolar Ca(OH)2 (74 g/mol), giving 0.0888 g/L. For 100 kg water (100 L), mass = 8.88 g ≈ 8.9 g.
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