To calculate the amount of H2 evolved from the complete reaction of 27g of aluminum with excess aqueous NaOH, we need to follow the steps below:



The balanced chemical equation for the reaction between aluminum and aqueous NaOH can be written as follows:

2Al(s) + 2NaOH(aq) + 6H2O(l) → 2Na[Al(OH)4](aq) + 3H2(g)



Since we have excess aqueous NaOH, we need to calculate the amount of NaOH used in the reaction. The molar mass of NaOH is 40 g/mol, so the number of moles of NaOH used can be calculated as follows:

Number of moles of NaOH = mass of NaOH used / molar mass of NaOH
Assuming excess NaOH, the mass of NaOH used is equal to the mass of aluminum used, which is 27g. Therefore,

Number of moles of NaOH = 27g / 40 g/mol = 0.675 mol



From the balanced chemical equation, we can see that 2 moles of aluminum react to produce 3 moles of H2. Therefore, the number of moles of H2 evolved can be calculated as follows:

Number of moles of H2 = (3/2) x number of moles of Al used
Number of moles of Al used = (27g / 27 g/mol) = 1 mol
Therefore,

Number of moles of H2 = (3/2) x 1 mol = 1.5 mol



At STP (standard temperature and pressure), one mole of any gas occupies 22.4 L. Therefore, the volume of H2 evolved can be calculated as follows:

Volume of H2 evolved = number of moles of H2 x molar volume at STP
Volume of H2 evolved = 1.5 mol x 22.4 L/mol = 33.6 L

Therefore, the amount of H2 evolved at STP on complete reaction of 27g of aluminum with excess aqueous NaOH would be 33.6 L.

2Al+ 2 NaoH+ 2H2o---->2NaAlo2+3H2


 2× 27 g of Al the volume of H2  liberated = 3 × 22.4 L (since 1 mole of H2 has 22.4 L at STP)so 2.7 g of Al  needs = 3 × 22.42 × 27 × 2.7 =3.36 L of H2 is  evolved.