Question

A 0.2755 g sample of an iron-carbon alloy was treated with 12.0 mL of a HCl solution. The alloy sample produced 74.5 mL of H2 gas at 2 atm and 25.0 oC. Determine the molarity of the HCl solution. Determine the mass % of iron in the alloy. 2 Fe + 6 HCl T 2 FeCl3 + 3 H2

Answer 1

2 Fe + 6 HCl--->FeCl3 + 3 H2

m = 0.2755

PV = nRT

n = PV/(RT) = 2*74.5/1000 / (0.082*298) = 0.0060975609 mol of H2

find moles of Hcl

since ratio is 3:6

then

0.0060975609*2 = 0.0121951218mol of HCl

then

M = mol/V = 0.0121951218/(12*10^-3) =  1.0162 M of HCl

c)

% mass of Fe

find moles of Fe

since ratio is 3:2 then 2/3*0.0060975609 = 0.0040650406 mol of Fe

mass = mol*MW = 0.0040650406*55.5 = 0.2256097533 g of Fe

% mass = Fe / total * 100 = 0.2256097533/0.2755 *100 = 81.8910% of Fe