Please help!!! Complete possible level of reactions' ('total and ionic', 'oxidation-reduction'): (12)
a) MgCl 2 + NaOH
b) Mg (NO3) 2+ CaCl2
c) Fe + CuCl2
d) CaCO3 + Al
1) Construct the full ionic and total reaction equation corresponding to this abbreviated ionic equation
Al3 ++ 3OH- → Al (OH) 3
2) How much precipitate will be formed by mixing solutions of 25 g of Fe2 (SO4) 3 and NaOH each?

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2