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

ID: 699198 • Letter: 6

Question

6) Hydrogen gas can be generated from the reaction between aluminum metal and hydrochloric acid: 2 Al (s) + 6HCI (aq) 2 AIC, (aq) + 3H2 (g) a. 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)? 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? b. Suppose the balloon is released and rises up to an altitude where the temperature is 11.2 °C and the pressure is 438 mm Hg. What is the new volume of the balloon (in L)? c. Page 3 of 3

Explanation / Answer

2Al(s) + 6HCl(aq) -----> 2AlCl3(aq) + 3H2(g)
2 moles of Al react with excess of HCl to gives 3 moles of H2
2*27g of Al react with excess of HCl to gives 3 moles of H2
3 g of Al react with excess of HCl to gives = 3moles*3g/2*27g
                                            = 0.166 moles of H2
n = 0.166 moles
V = 850ml = 0.85L
T = 24+273 = 297K
PV = nRT
P   = nRT/V
    = 0.166*0.0821*297/0.85 = 4.76atm
b.

experimetal conditions                              STP conditions
P1 = 4.76atm                                         P2 = 1atm
V1 = 0.85L                                           V2 =
T1 = 297K                                             T2 = 273K
         P1V1/T1   = P2V2/T2
            V2     = P1V1T2/T1P2
                   = 4.76*0.85*273/297*1
                   = 3.72L >>>>.answer at STP conditions
c.
STP conditions                                new conditions
P1 = 1atm = 760 mmHg                          P2 = 438mmHg
V1 = 3.72L                                    V2 =
T1   = 273K                                     T2 = 11.2+273 = 284.2K
                P1V1/T1   =     P2V2/T2
                   V2     = P1V1T2/T1P2
                          = 760*3.72*284.2/273*760   = 3.87L >>>>answer

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