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Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)What mass of CaCO3 is required to completely with 25 mL of 0.75 MHCl? |
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Answer» CaCO3 (s) + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l) 1 mol 2 mol (40 + 12 + 48) g 2(1 + 35.5) g = 100 g = 73 g Amount of HCl in 25 mL of 0.75 M HCl = M x V = 0.75 mpl L-1 x 25 mL = 0.75 mol L-1 x 25/1000 L = {0.75 x 25}/{1000} mol = 0.0188 mol 2 mol of HCl reacts with 100 g CaCO3 0.0188 mol of HCl reacts with = {100 g x 0.0188}/{2} g CaCO3 = 0.94 g CaCO3 |
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