3) Calculate the cell potential of the following electrochemical cell: CroCr3+(0.0250M) || Ag+ (aq) (0.0400M) | Age)
Standard Reduction Potentials
E
Cr3+(aq) + 3e → Cr(s)
-0.74 V
Ag (aq) + e → Ag(s)
+0.80 V

the answer is c. nacl

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