Question

Hydrobromic acid (HBr) dissolves solid iron according to the reaction: $$\mathrm{Fe}(s)+2 \mathrm{HBr}(a q) \longrightarrow \mathrm{FeBr}_{2}(a q)+\mathrm{H}_{2}(\mathrm{g})$$ What mass of \(\mathrm{HBr}(\) in \(\mathrm{g})\) do you need to dissolve a 3.2 -g purc iron bar on a padlock? What mass of \(\mathrm{H}_{2}\) can the complete reaction of the iron bar produce?

Short answer

To dissolve a 3.2 g iron bar, you would need 11.46 g of HBr and the reaction would produce 0.057 g of H2.

Step-by-step solution

Calculate the moles of iron (Fe)

Using the molar mass of iron (55.85 g/mol), calculate the amount of moles of Fe in a 3.2 g iron bar. Moles of Fe = mass of Fe / molar mass of Fe.

Determine the moles of HBr required

According to the balanced reaction, 2 moles of HBr react with 1 mole of Fe. Thus, multiply the moles of Fe by 2 to find the moles of HBr required.

Calculate the mass of HBr needed

Using the molar mass of HBr (80.91 g/mol), calculate the mass of HBr needed by multiplying the moles of HBr by the molar mass of HBr.

Determine the moles of H2 produced

The balanced reaction indicates that 1 mole of Fe produces 1 mole of H2. Therefore, the moles of H2 produced will be equal to the moles of Fe.

Calculate the mass of H2 produced

Using the molar mass of H2 (2.016 g/mol), calculate the mass of H2 produced by multiplying the moles of H2 by the molar mass of H2.

Key concepts

Molar Mass Calculation

Understanding molar mass is critical when working with chemical reactions. The molar mass, expressed in grams per mole (\text{g/mol}), is the weight of one mole of a substance. One mole is equivalent to Avogadro's number, approximately \(6.022 \times 10^{23}\) entities, be it atoms, molecules, ions, or other particles.

To calculate the molar mass, you need the periodic table to find the atomic mass of each element in the compound. For example, iron (Fe) has an atomic mass of about 55.85, so its molar mass is \(55.85 \text{g/mol}\). Hydrobromic acid (HBr) comprises hydrogen (H) and bromine (Br), with atomic masses of approximately 1 and 79.9, respectively. Thus, the molar mass of HBr is the sum of the atomic masses of hydrogen and bromine, which equals \(1 + 79.9 = 80.9 \text{g/mol}\).

When calculating the molar mass in a chemical equation, ensure each component is multiplied by its coefficient in the balanced equation to maintain stoichiometric ratios.

Chemical Reaction Balancing

Balancing a chemical reaction is vital for stoichiometry. It ensures that the number of atoms for each element is the same on both the reactant and product sides of a chemical equation. This reflects the conservation of mass principle: matter is neither created nor destroyed during a chemical reaction.

In the reaction between hydrobromic acid (HBr) and iron (Fe) to form iron(II) bromide (\text{FeBr}_2) and hydrogen gas (\text{H}_2), the balanced chemical equation is:\[ \text{Fe}(s) + 2 \text{HBr}(aq) \to \text{FeBr}_2(aq) + \text{H}_2(g) \]This indicates that one mole of iron reacts with two moles of hydrobromic acid to produce one mole of iron(II) bromide and one mole of hydrogen gas. The coefficients (1 for Fe and \text{FeBr}_2 and 2 for HBr and \text{H}_2) must be correctly used when calculating the moles of each substance involved in the reaction.

Moles to Mass Conversion

Once the number of moles in a chemical reaction is understood, moles to mass conversion becomes a straightforward multiplication by the molar mass of the substance. This conversion is crucial in preparing compounds and solutions for reactions and in determining yields.

For instance, if you have the number of moles of hydrobromic acid needed to react with iron, you can convert this to grams by multiplying by the molar mass of hydrobromic acid. The formula looks like this:\[ \text{Mass (g)} = \text{Moles} \times \text{Molar Mass (g/mol)} \]
In the lock and key reaction—where a solid iron bar is dissolved by hydrobromic acid—the mass of the hydrogen gas produced in the reaction can also be calculated in a similar manner. Knowing the moles of iron and HBr involved, we utilize their respective molar masses (for \text{H}_2, this is \(2.016 \text{g/mol}\)) to find the total grams of each substance through simple multiplication.