What volume, in mL, of 0.7376 molar hydrochloric acid is requiredto react with 1.1835 g of calcium carbonate? Assume a 100% yieldand report your answer to the correct number of significantdigits. DONOT enter units as part of your answer.

Useful information: C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00g/mol, Ca = 40.08 g/mol, Cl = 35.45 g/mol

2 HCl (aq)+ CaCO₃ (aq)---> CaCl₂ (aq)+ CO₂ (g)+ H₂O(l)

Answer rating: 100% (QA)

At equilibrium moles of CaCO3 2 moles of ... View the full answer