The equation for the reaction between Al(s) and Pb+2(aq) is: 2Al(s) + 3Pb+2(aq) 2Al+3(aq) + 3Pb(s) You know that 0.011 moles of Al(s) have been consumed and that 0.017 moles of Pb(s) have been formed. Calculate [Pb+2].

1 is the answer to this problem

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