Question

the steps needed to the answer?

9. What volume of a 0.200 M hydrochloric acid (36.46 g/mol) solution is required to completely react with 0.550 g of aluminum hydroxide (78.01 g/mol)? Al(OH)3(s) ЗНСКад) AICI3(aq) + 3H2O(I a. 35.3 mL b) 106 mL c. 0.106 mL d. 11.8 mL

Answer 1

Al 6H), (0) + 3 Hel lay) Alley (ag) + 3 H20 (1) mol- men mannen moles of AL (OH)₂ = 0.5509 78.01g/mol - 705 x 103 mol 3x7.05x103 nol moly of HQ & 0.0 211 2 rele Allery- moly mobarily = 0.02112 mel 0.2.00 mol/L o.lost ml - 105.7 me - 106 ml 61