6) Hydrogen gas can be generated from the reaction between aluminum metal and hy

Question

6) Hydrogen gas can be generated from the reaction between aluminum metal and hydrochloric acid: 2 Al (s) + 6 HCl (aq) 2 Alci, (aq) + 3H2(g) Suppose that 3.00 grams of Al are mixed with excess acid. If the hydrogen gas produced is directly collected into a 850. mL glass flask at 24.0 °C, what is the pressure inside the flask (in atm)? a. b. This hydrogen gas is then completely transferred from the flask to a balloon. To what volume (in L) will the balloon inflate under STP conditions? Suppose the balloon is released and rises up to an altitude where the temperature is 11.2 °C and the pressure i 438 mm Hg. What is the new volume of the balloon (in L)? c. Page 3 o 109

Explanation / Answer

6)

(a) Molar mass of Al = 26.98 g/mol
No. of moles of Al reacted = (3.00 g) / (26.98g/mol) = 0.11 mol

2Al(s) + 6HCl(aq) ----> 2AlCl3(aq) + 3H2(g)
Mole ratio of Al : H2 = 2 : 3
Number of moles of H2 formed = (0.11 mol Al) x (3 mol H2 /2 mol Al) = 0.165 mol

Now, for the hydrogen gas formed ,
P = ?  
V = 850 mL = 0.850 L
n = 0.165 mol
R = 0.08206 L atm / mol K
T = (24.0 + 273 ) K = 297 K

PV = nRT
=>P = nRT/V
Pressure of H2 gas, P = 0.165 mol x 0.08206 Latm/molK x 297 K / 0.850 L

=> P= 4.73 atm

(b)  P = 1 atm ( at STP)
V = ?  
n = 0.165
R = 0.08206 L.atm/mol K
T = 273 K ( at STP)

PV = nRT
=>V = nRT/P

Volume of H2 gas, V = 0.165 mol x 0.08206 L atm/mol K x 273 K / 1 atm

=> V= 3.69 L

(c)  P = 438 mmHg
V = ?  
n = 0.165 mol
R = 62.36 L mmHg / mol K
T = (11.2 + 273) K = 284.2 K

PV = nRT
=>V = nRT/P

Volume of H gas, V = 0.165 mol x 62.36 L mmHg/mol K x 284.2 K / 438 mmHg

=> V = 6.67 L

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