This problem requires knowledge of stoichiometry and the molar mass of iron.
First, let's write the balanced chemical equation for the reaction between iron and sulfuric acid:
Fe + H2SO4 -> FeSO4 + H2
From this equation, we can see that 1 mole of iron reacts with 1 mole of sulfuric acid to produce 1 mole of hydrogen gas. Therefore, we can use the amount of hydrogen produced to determine the amount of iron that reacted:
n(H2) = PV/RT = (1 atm)(0.112 L)/(0.0821 L atm/mol K)(298 K) = 0.00455 moles H2
Since the mole ratio of Fe to H2 is 1:1, we know that 0.00455 moles of iron reacted with the sulfuric acid. Now we need to convert this to mass:
mass(Fe) = n(Fe) x M(Fe)
We don't know the molar mass of the impure iron rod, but we can assume that it is mostly iron and use the molar mass of pure iron (55.85 g/mol) as a reasonable estimate. We can then use the mass of the entire rod, given in the problem, to calculate the mass of pure iron:
mass(pure Fe) = (55.85 g/mol)(0.25 kg) = 14 g
Now we can calculate the number of moles of pure iron that reacted:
n(pure Fe) = mass(pure Fe)/M(pure Fe) = 14 g/55.85 g/mol = 0.25 moles
Since the mole ratio of Fe to H2 is 1:1, we know that 0.25 moles of pure iron reacted with the sulfuric acid. Finally, we can use this to calculate the mass of pure iron that reacted:
mass(pure iron reacted) = n(pure Fe) x M(pure Fe) = 0.25 moles x 55.85 g/mol = 13.96 g
Therefore, approximately 14 grams of pure iron reacted with the sulfuric acid to produce 1.12 dm³ of hydrogen gas.
4g if impure iron rod was use to stir H2So4 and 1.12dm of H2 was liberated determine the Mass of pure iron rod that reacted with H2SO4
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