Problem # 1: (50) In a laboratory react 5.0 moles of aluminum metal with hydroch

Question

Problem # 1: (50) In a laboratory react 5.0 moles of aluminum metal with hydrochloric acid containing 17 mol of acid. A time after the reaction, the solution resulting has a volume of 2228.0 mL and a total mass of 2666.8 g. a) Write the reaction that take place in the process. (2) b) Determine the limiting reactant and the excess reactant. (4) e) Calculate the amount of substance (moles) of aluminum chloride obtained in the process. (3) d) Calculate the mass of salt obtained. (3) e) Calculate the amount of the gas generated. (3) f Calculate the amount of substance that not reacted. (3) g) Calculate the volume of the gas formed at STP. (3) h) If the solution of aluminum chloride is the desired product, calculate the yield percent. (3) ) Calculate the concentration of aluminum chloride in the solution (Molarity). (3) ) Calculate the molality of aluminum chloride in the final solution. (2) k) Calculate the weight percent of each substance in the final solution. (3) l) Determine the average molecular weight of the final solution. (2) m) Calculate the density of the final solution. (2) n) Calculate the mass selectivity of the process. (2) o) Give a sketch of the process. (3) p) Calculate the molar and mass fractions of each stream. (4) q) Calculate the conversion of each reactant. (2) r) Calculate the percent of excess in the process. (3)

Explanation / Answer

Part a

The balanced reaction

2 AL(s) + 6 HCl(aq) = 2 ALCl3(aq) + 3 H2(g)

Aluminum + Hydrochloric acid = Aluminum chloride + hydrogen

Part b

Moles of Al = 5

Moles of HCl = 17

From the stoichiometry of the reaction

2 mol of Al reacts with = 6 moles of HCl

5 mol of Al reacts with = 6*5/2 = 15 moles of HCl

We have 17 moles of HCl which is in excess.

HCl = excess reactant

Al = limiting reactant

Part C

Moles of ALCl3 obtained = 2 mol AlCl3 x 5 mol Al / 2 mol Al

= 5 mol AlCl3

Part d

Mass of AlCl3 = moles x molecular weight

= 5 mol x 133.34 g/mol

= 666.7 g

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