compute for the volume of Hydrogen produced at 55°C and 1.5 atm when 3.5 g of AI is used up in the reaction:
2 AI(s)+6HCI (aq) ——2 AICI2 (aq) + 3H2(g)
2AI(s) + 6HCI(aq) —> 2AICI2(aq) + 3H2(g)
n(Al) = m(Al) / Mr(Al) = 3.5 g / 27.0 g/mol = 0.13 mol
2 mol Al - 3 mol H2
0.13 mol Al - x mol H2
x = 0.13 × 3 / 2 = 0.195 mol
pV = nRT
V = nRT/p = (0.195 mol)(0.0821 atm L/mol K)(55 + 273 K) / (1.5 atm) = 3.5 L
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