How many grams of calcium carbonate are needed to completely react with 6.22g of hydrochloric acid (HCI) in the equation below? Note: respond to three significant figures. 1CaCO3(s) + 2HCl(aq) --> 1CaCl2(s) + 1CO2(g) + 1H20(1) + #1 #2 #3 #4 #5 #6 #7 #7 Answer #8: Word Bank: 4.53 g HC 4.53 g CaCo3 1 mole CaCO3 100.09 g CaCO3 216.875 CaCo3 6.22 g HCI 36.458 g HCl 2 moles HCI 1 mole CaCo3 1 mole HCI 8.54 g CaCO3 2 mole CaCO3