Limestone (CaCO3) is commonly identified using the reaction below. If a 2.50 g s

Question

Limestone (CaCO3) is commonly identified using the reaction below. If a 2.50 g sample of CaCO3 reacts with 2.50 mL of 0.100 M HCl, what will be the limiting reactant, and how many moles of the excess reactant will be left over? CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

The reaction below is used to standardize a constant pressure calorimeter. At an initial temperature of 25.00C, 0.500 g of C4H11NO3 is mixed with 100.0 mL of 0.1000 M HCl. 123 J of heat is then released into the water. Calculate the final water temperature, assuming swater = 4.184 J/gC and dwater = 1.00 g/mL. C4H11NO3(aq) + HCl(aq) C4H12NO3Cl(aq)

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Explanation / Answer

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

No.of moles of CaCO3 = 2.50 / 100 = 0.025 mol

No. of moles of HCl = 0.1 x 2.5 / 1000 = 0.00025 mol

No. of moles of HCl required for 0.025 mol of lime stone = 0.025 x 2 = 0.50 mol

Hence the limiting reagent = HCl

Left over amount of excess reagent i.e lime stone = 0.025 - (0.00025/2) = 0.024875 mol

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