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How many grams of CaCl2 are formed when 15.00 mL of 0.00237 mol L-1 Ca(OH) 2 reacts with excess Cl2 gas?
2Ca(OH) 2(aq) + 2Cl2(g) → Ca(OCl) 2(aq) + CaCl2(s) + 2H2O(l)
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