2 2. A sample of iron reacted with hydrochloric acid. The liberated hydrogen hyd

Question

2 2. A sample of iron reacted with hydrochloric acid. The liberated hydrogen hydrochl occupied 40.1 ml when the aese of the collea barometric pressure of 750 torr. What is the mass of the sample of must be subtracted arometric on of gas over water, the vapor pressure of the water pressure to find the pressure of the hydrogen gas.) ncial you wanted to prepare 50.0 mL of hydrogen, collected over water at 25 C on a day when the arometric pressure was 730 torr, what mass of aluminum would you react with a hydrochloric acid? 3. If (Balance the equation first.) H2 (g) + ? Cls (aq) HCI (aq) Al (s)

Explanation / Answer

2)

2Fe(s) + 6HCl(aq) ---> 2FeCl3(s) + 3H2(g)

2 mol Fe(s) = 6 mol HCl(aq) = 3 mol H2(g)

No of mol of H2 liberated(n) = PV/RT

   P = pressure of H2-gas = Pobs - Pwater

                          = 750 - 26.7

           = 723.3 torr

                         = 0.952 atm

V = 40.1 ML = 40.1*10^-3 L

R = 0.0821 l.atm.k-1.mol-1

T = 27 C = 300 k

n = (0.952*40.1*10^-3)/(0.0821*300)

    = 0.00155 mol

No of mol of Fe reacted = 0.00155*2/3 = 0.001 mol

mass of Fe reacted = n*Mwt

                    = 0.001*56

                    = 0.056 g

3) no of mol of H2(n) = PV/RT

P = pressure of H2-gas = Pobs - Pwater

                          = 730 - 23.8

           = 706.2 torr

                         = 0.93 atm

V = 50 ML = 50*10^-3 L

R = 0.0821 l.atm.k-1.mol-1

T = 25 C = 298 k

n = (0.93*50*10^-3)/(0.0821*298)

    = 0.0019 mol

balanced equation : 2Al + 6HCl ---> 2AlCl3 + 3H2

2 mol Al = 3 mol H2

No of mol of Al required = 0.0019*2/3 = 0.001267 mol

mass of Al required = 0.001267*27 = 0.0342 g

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