When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2al(s)+6hcl(aq)-> 2alcl3(aq) +3h2(g) what mass of al(s) is required to produce 513.0 ml of h2(g) at stp?

Answer : The mass of aluminium (Al) is 0.413 g

Solution : Given,

Volume of at STP = 513 ml = 0.513 L           ( 1 L = 1000 ml )

Molar mass of aluminium = 26.98 g/mole

First we have to calculate the moles of .

At STP,    

1 mole occupies 22.4 L volume

now, 0.513 L gives moles of

The moles of = 0.0229 moles

The Net balanced chemical reaction is,

From the balanced chemical reaction, we conclude that

2 moles of Aluminium (Al) produces 3 moles of hydrogen gas

Now the number of moles of aluminium required in 0.0229 moles of hydrogen gas = = 0.0153 moles

Now we have to calculate the mass of aluminium.

Mass of aluminium = number of moles × Molar mass = 0.0153 moles × 26.98 g/mole = 0.413 g

The mass of aluminium required is 0.413 g.