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Thermochemistry/gas laws 2. Aluminum metal reacts with hydrochloric acid to prod

ID: 697765 • Letter: T

Question

Thermochemistry/gas laws 2. Aluminum metal reacts with hydrochloric acid to produce hydrogen gas. A. If excess HCl reacts with 2.70g Aluminum how many moles of hydrogen will be collected. B. If the dry hydrogen gas is collected at 28.0°C and 750mm of Hg, what volume will be collected? The H(kJ/mol) for that reaction is-1500kJ/mole, how much heat would be released when the 2.7g of aluminum reacts? C. D. If the volume of water surrounding the reaction is 100mL (Assume 1.00g/mL) what would be the temperature change of the water if no heat escapes E. What is the sign of q (heat) and w (work) for this reaction? Explain.

Explanation / Answer

    2 mol Al = 6 mol HCl = 2 mol AlCl3 = 3 mol H2

a) no of mol of Al = w/mwt = 2.7/27 = 0.1 mol

no of mol of Al COLLECTED = 0.1*3/2 = 0.15 mol

b) volume of gas(V) = nRT/P

                    = 0.15*0.0821*(28+273.15)/(750/760)

                    = 3.76 L

c) heat released(q) = DH*n

                    = -1500*0.1/2

                   = -75 kj

d) q = m*s*DT

75*10^3 = 100*4.184*x

x = change in temperature = DT = 179.25 C

e) heat is released(q) = - ve,

   work done by the system (w) = -ve

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