Answer:
V = 11.2L are produced
Explanation:
... Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..
Based on the chemical reaction:
2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)
Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.
To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:
Moles Al-Molar mass: 26.98g/mol-:
27g * (1mol / 26.98g) = 1mol of Al
Moles HCl:
333mL = 0.333L * (3mol/L) = 1mol HCl
For a complete reaction of 1 mole of HCl are required:
1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and HCl is limiting reactant.
Moles of Hydrogen produced are:
1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.
Using ideal gas law:
PV = nRT
V = nRT/P
Where V is volume
n are moles: 0.5mol
R is gas constant: 0.082atmL/molK
T is absolute temperature: 273.15K
P is pressure: 1atm.
Solving for V:
V = 0.5mol*0.082atmL/molK*273.15K / 1atm
