Question

How many mL of 0.758 M HCl are needed to dissolve 5.79 g of CaCO,? 2HCl(aq) + CaCO3(s) CaCl(aq) + H2O(l) + CO2(8) mL Submit Answer Retry Entire Group 9 more group attempts remaining sited Previous Nerd Save al

Answer 1

2HCl (48) + Caco, (s) —> CaCl, 169) + H2O(l) + CO. (8) 2 mole I mole 2x 36.5 g = 100g = 73 g can from balanced 100 g CaCO₃ reaid with = 73 g HCl & 5.79 g CaCO will reach with = 73 *8.79 g Hel 100 14.2267 g Hl. Now molay mass of HCl = 36.5 g/mole 0.758 molar solution means, 0.758 mole HCl are dissolved in = 1000 mL or 27.667 g HCl are dissolved in = 1000 mL & 4.2267 g HCl will be dissolved in a

ml 4.2267 x 1000 27.667 = 1152.77 mL Answer Note: - 36.5 g Hel = I mole or I mole Hel = 36.5g 0.750 mole HCl = 36:50 0.750 g = 27.667 g