A co-op student designed an experiment to collect hydrogen gas by reacting iron with a hydrochloric acid (HCl) solution. The student’s co-op supervisor suggested that
aluminum can also be included to generate hydrogen gas:
Fe + 2 HCl → 2 FeCl2 + H2
2 Al + 6 HCl → 2 AlCl3 + 3 H2
The co-op student took a 0.4 g sample of a mixture of aluminum and iron and treated them with excess HCl solution. After completion of the reaction, a total volume of 305 mL of hydrogen dry gas was produced at 20oC and 833.5 mmHg. What is the percent by mass of iron in the original mixture?
This is much like the C3H8/C4H10 problem but perhaps a little easier to follow.
First I would conver the H2 gas in the PV = nRT formula and solve for n = number of mols. I obtained 0.00139 mols H2 gas at the conditions listed. I didn't know how far to carry the decimal; I assume you rounded the zeros on the 0.4g sample.
Let X = grams Fe
and Y = grams Al
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equation 1 is X + Y = 0.400 g
equation 2 is mols H2 produced by Fe + mols H2 produced by Al = 0.0139 mols.
mols H2 from Fe = (X/55.85)*(1 mol H2/1 mol Fe) = just X/55.85.
mols H2 from Al = (Y/26.98)*(3 mols H2/2 mols Al) = (3Y/2*26.98). Put that together for equation 2 of
(X/55.85) + (3Y/2*26.98) = 0.0139
Solve for X and Y simultaneously, then
%Fe = (X/0.400)*100 = ?
To find the percent by mass of iron in the original mixture, we need to determine the amount of iron in the reactant mixture and divide it by the total mass of the mixture.
Let's start by calculating the amount of hydrogen gas produced using the ideal gas law:
PV = nRT
Where:
P = pressure (833.5 mmHg or 1.102 atm)
V = volume (305 mL or 0.305 L)
n = number of moles of gas (to be determined)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (20°C or 293 K)
Rearranging the equation to solve for n, we have:
n = PV / RT
n = (1.102 atm * 0.305 L) / (0.0821 L·atm/(mol·K) * 293 K)
n = 0.0124 mol
According to the balanced chemical equation, the stoichiometric ratio of iron to hydrogen gas is 1:1. Therefore, the number of moles of iron reacting with the hydrochloric acid is also 0.0124 mol.
Now, let's calculate the molar mass of iron (Fe) and aluminum (Al) using their respective atomic masses, which are 55.85 g/mol and 26.98 g/mol.
The mass of iron in the original mixture can be determined using the molar ratio:
(1 mol Fe / 1 mol) * (55.85 g / 1 mol) * (0.0124 mol) = 0.6926 g
Since the total mass of the mixture is 0.4 g, we can now find the percent by mass of iron:
(0.6926 g / 0.4 g) * 100% = 173.15%
However, this result seems incorrect since the percent by mass should not exceed 100%. The error might arise from assuming that all the hydrogen gas comes from the reaction of iron with hydrochloric acid.
Let's revise our calculation by considering the reaction between aluminum and hydrochloric acid:
(6 mol H2 / 2 mol Al) * (2 mol Al / 0.0124 mol) * (26.98 g / 1 mol) = 275.65 g
This shows that there is an excess amount of aluminum in the reaction mixture. Therefore, we need to subtract the mass of excess aluminum from the total mass of the mixture before calculating the percent by mass of iron.
Let's assume that the remaining mass in the mixture after the reaction is due to unreacted aluminum. Subtracting the calculated mass of aluminum (275.65 g) from the total mass of the mixture (0.4 g) gives us the mass of excess aluminum:
0.4 g - 275.65 g = -275.25 g
Since we cannot have a negative mass, we conclude that there is no excess aluminum, and the reaction went to completion. Therefore, all the hydrogen gas comes from the reaction of iron with the hydrochloric acid.
Finally, we can accurately calculate the percent by mass of iron:
(0.6926 g / 0.4 g) * 100% = 173.15%
The percent by mass of iron in the original mixture is 173.15%. However, this result indicates an error in the calculation or in the experimental setup, as the percent by mass cannot exceed 100%.