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When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2Al(s)+6HCl(aq)⟶2AlCl3(aq)+3H2(g) What volume of H2(g) is produced when 7.40 g Al(s) reacts at STP?

Sagot :

Answer:

9.18 L

Explanation:

When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2Al(s)+6HCl(aq)⟶2AlCl3(aq)+3H2(g) What volume of H2(g) is produced when 7.40 g Al(s) reacts at STP?

first write, then balance the equation

Al + HCl-------------> AlCl3 + H2

2Al + 6HCl-----------> 2AlCl3 + 3H2 2 MOLE  

2 moles of Al produce 3 moles of H2

Al is 7.4/27 =

0.27 moles of Al.

this will produce

0.27 X 3/2 moles of H2 =

0.41 moles of H2

PV =nRT

for STP, P = 1 atm, T = 273.2, R IS 0.082

V= (0.41) X 0.082 X 273,2/1 = 9.18 L

CHECK 1 mole of gas at STP has a volume of 22.4 L

0.41 moles then has a volume of 00.41 X 22.4L = 9.,18 L

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