How many liters of 0.85 m hcl solution would react completely with 3.5 moles ca(oh)2? ca(oh)2(s) + 2 hcl(aq) yields cacl2(aq) + 2 h2o(l)

Ca(OH)2(s) + 2 HCl(aq) > CaCl2(aq) + 2 H2O(l)
3.5 moles        2*3.5 moles

we need  2*3.5 moles HCl= 7.0 mol HCl
7 mol HCl *1L solution HCl/0.85 mpl HCl=8.2 L 0.85 M HCl solution
8.2 L 0.85 M HCl solution

Cis the answer ur welcome if i get it wrong u don't have to me

Remember that rust prevented by placing a sacrificial electrode in electrical contact with the iron. the sacrificial electrode must be composed of a metal that  oxidizes  more easily than iron (that is, it must be below iron in table 18.1).