If 12.0grams of aluminum reacts with HCL accorinding to the reaction below how many liters of hydrogen gas...?
Answer:
the balanced equation is
2Al + 6 HCl >>2 AlCl3 +3 H2
moles Al = 12.0 /26.98 = 0.445
the ratio between Al and H2 is 2 : 3
2 : 3 = 0.445 : x
x = 0.667 moles H2 produced
V = nRT / p
at STP
p = 1 atm and T = 273 K
V = 0.667 x 0.0821 x 273 / 1 = 14.9 L of H2
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