the calcium carbonate in limestone reacts with a HCl solution to produce a calci

Question

the calcium carbonate in limestone reacts with a HCl solution to produce a calcium chloride solution, carbon dioxide, and water. CaCO3(s) + 2HCl(aq) > CaCl2(aq) +H2O(l) +CO2(g). a) how many milliliters of a .200 M HCl solution can react with 8.25 g of CaCO3? b) how many liters of CO2 gas can form at STP when 15.5 mL of a 3.00 M HCl solution reacts with excess CaCO3? c)what is the molarity of a HCl solution if the reaction of 200 mL of the HCl solution with excess CaCO3 produces 12.0 L of CO2 gas at 725 mmHg and 18 degrees C

Explanation / Answer

Molecular Weights of HCl = 36.5 gm/mol CaCO3 = 100 gm/mol CO2 = 44 gm/mol CaCl2 = 111 gm/mol H2O = 18 gm/mol Solution : (a) by stoichoimetry, 1 mol CaCO3 reacts with 2mol HCl we have 8.25 gm of CaCO3 or 8.25/100 = 0.0825mol CaCO3 for this we need 2*0.0825 = 0.165 mol HCl Molarity of HCl = 0.2M so volume of 0.2M HCl reqd = 0.165/0.2 = 0.825liter or 825ml of HCl (b) we have 15.5ml of 3M HCl or (15.5*3)/1000 = 0.0465 mol HCl as CaCO3 is in excess, HCl is limiting reagent and CO2 evolved will be proportional of HCl reacted assume complete consumption of HCl and ideal behavior of CO2 by stoichiometry, mol CO2 formed = 0.5*mol HCl reacted = 0.02325 mol at STP 1 mol of an ideal gas occupies 22.4 liters of volume so 0.02325 mol CO2 will occupy 0.5208 liters at STP (c) Using ideal gas eqn, mol CO2 formed = (PV/RT) P = 725mm Hg = (725/760) atm = 0.954 atm V = 12L T = 18 deg C = (18+273)K = 291K R = 0.08206 L.atm./(mol.K) put all values, mol CO2 formed = 0.4794 mol CO2 which means 2*0.4794 mol HCl reacted (limiting reagent is HCl, as CaCO3 is in excess) mol HCl = 0.959mol volume of HCl = 200ml =0.2L hence Molarity of this HCl = 0.959/0.2 = 4.794M

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