Answer:
[tex]\large \boxed{\text{46.1 L}}[/tex]
Explanation:
We will need a balanced chemical equation with molar masses, so, let's gather all the information in one place.
Mᵣ: 55.84
2Fe + 6HCl ⟶ 2FeCl₃ + 3H₂
m/g: 70.1
(a) Moles of Fe
[tex]\text{Moles of Fe } =\text{70.1 g Fe } \times \dfrac{\text{1 mol Fe }}{\text{55.84 g Fe }} =\text{1.255 mol Fe}[/tex]
(b) Moles of H₂
The molar ratio is 3 mol H₂:2 mol Fe
[tex]\text{Moles of H$_{2}$}= \text{1.255 mol Fe} \times \dfrac{\text{3 mol H$_{2}$}}{ \text{2 mol Fe}} = \text{1.883 mol H$_{2}$}[/tex]
(c) Volume of H₂
We can use the Ideal Gas Law to calculate the volume of hydrogen.
pV = nRT
[tex]\rm V = \dfrac{nRT}{p}= \dfrac{\text{1.883 mol $\times$ 0.08206 L$\cdot$atm$\cdot$K$^{-1}$mol$^{-1}\times$ 298.15 K}}{\text{ 1 atm}} = \textbf{46.1 L} \\\\\text{You can produce $\large \boxed{\textbf{46.1 L}} $ of H$_{2}$.}[/tex]
