A 2.05g sample of an iron-aluminum alloy (ferroaluminum) isdissolved in excess H

Question

A 2.05g sample of an iron-aluminum alloy (ferroaluminum) isdissolved in excess HCL(aq) to produce 0.105g H2(g).What is the percent composition, by mass of theferroaluminim? Fe(s) + 2 HCl(aq) -> FeCl2(aq) +H2(g) 2 Al(s) + 6 HCl(aq) -> 2 AlCl3(aq) + 3H2(g) Please include work and any helpful explainations.Thanks! A 2.05g sample of an iron-aluminum alloy (ferroaluminum) isdissolved in excess HCL(aq) to produce 0.105g H2(g).What is the percent composition, by mass of theferroaluminim? Fe(s) + 2 HCl(aq) -> FeCl2(aq) +H2(g) 2 Al(s) + 6 HCl(aq) -> 2 AlCl3(aq) + 3H2(g) Please include work and any helpful explainations.Thanks!

Explanation / Answer

Let the mass of iron be x and that of aluminium be y x + y = 2.05 No. of moles of Hydrogen formed = 0.105/2 Now this is equal to the no. of moles formed from iron andaluminium according to the reactions given 0.105/2 = (x/56) + (y/27)*(3/2) Solving the two equations in x and y we get x = 1.63 g and y = 0.42 g Thus the mass percentage of Iron is = 1.63/2.05 = 80% And mass percentage of aluminium is = 0.42/2.05 = 20%

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