How many milliliters of 0.997 M HCl are needed to react with 46.7 g of CaCO3 ? 2HCl(aq) + CaCO3 (s) → CaCl2(aq) + CO2(g) + H2O(l)?
- Roger the MoleLv 73 weeks ago
(46.7 g CaCO3) / (100.0875 g CaCO3/mol) x (2 mol HCl / 1 mol CaCO3) /
(0.997 mol HCl/L) = 0.935991 L = 936 mL HCl
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