What minimum mass of HCl in grams would you need to dissolve a 2.5g iron bar on a padlock?

To determine the minimum mass of HCl needed to dissolve the iron bar, we need to calculate the number of moles of iron present in the bar and then use the stoichiometry of the reaction between iron and hydrochloric acid (HCl) to find the minimum moles of HCl required.

1. Calculate the number of moles of Fe using its molar mass:
Molar mass of Fe = 55.85 g/mol
Moles of Fe = Mass of Fe / Molar mass of Fe
= 2.5 g / 55.85 g/mol

2. Write and balance the chemical equation for the reaction between iron and hydrochloric acid:
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)

From the balanced equation, we see that 1 mole of Fe reacts with 2 moles of HCl.

3. Calculate the minimum moles of HCl needed:
Moles of HCl = Moles of Fe × (2 moles of HCl / 1 mole of Fe)

4. Convert the moles of HCl into grams using its molar mass:
Molar mass of HCl = 36.46 g/mol
Mass of HCl = Moles of HCl × Molar mass of HCl

By following these steps, you can obtain the minimum mass of HCl required to dissolve the iron bar on the padlock.

To find the minimum mass of HCl required to dissolve a 2.5g iron bar, we need to calculate the stoichiometric ratio between iron and hydrochloric acid.

The chemical equation for the reaction between iron and hydrochloric acid is:

Fe(s) + 2HCl(aq) -> FeCl2(aq) + H2(g)

From the equation, we can see that one mole of iron reacts with two moles of hydrochloric acid.

1 mole of Fe = 55.85 g
2 moles of HCl = (1 x 2) x (1.01 + 35.45) = 73.36 g

The molar mass of HCl is 36.46 g/mol (1.01 g/mol for hydrogen + 35.45 g/mol for chlorine).

Therefore, the stoichiometric ratio between iron and hydrochloric acid is:

55.85 g Fe : 73.36 g HCl

To dissolve the 2.5g iron bar, we can set up a proportion:

55.85 g Fe / 73.36 g HCl = 2.5 g Fe / x g HCl

Solving for x, we get:

x = (2.5 g Fe / 55.85 g Fe) * 73.36 g HCl
x ≈ 3.31 g HCl

So, you would need a minimum mass of approximately 3.31 grams of HCl to dissolve a 2.5g iron bar.

See this solved example.

http://www.jiskha.com/science/chemistry/stoichiometry.html