Respuesta :
Answer:
3.49 Liters of hydrogen is produced when 2.80 grams of aluminum reacts at STP.
Explanation:
[tex]2Al ( s ) + 6HCl ( aq )\rightarrow 2AlCl_3 ( aq) +3H_2 ( g )[/tex]
Moles of aluminum = [tex]\frac{2.80 g}{27 g/mol}=0.1037 mol[/tex]
According to reaction , 2 moles of aluminium gives 3 moles of hydrogen gas.Then 0.1037 moles of aluminum will give:
[tex]\frac{3}{2}\times 0.1037 mol=0.1556 mol[/tex] of hydrogen gas
Moles of hydrogen gas = n = 0.1556 mol
Pressure of the gas, at STP , P = 1 atm
Temperature of the gas, at STP = T = 273 K
Volume of the gas , At STP = V
[tex]PV=nRT[/tex] ( Ideal gas equation )
[tex]V=\frac{nRT}{P}=\frac{0.1556 mol\times 0.08210 atm L/mol K\times 273 K}{1 atm}=3.49 L[/tex]
3.49 Liters of hydrogen is produced when 2.80 grams of aluminum reacts at STP.
