Question

What volume of hydrogen gas is produced when 70.1 g of iron reacts completely according to the following reaction at 25 °C and 1 atm? iron (s) + hydrochloric acid(aq)iron(II) chloride (aq) + hydrogen(g) liters hydrogen gas

Answer 1

The volume of hydrogen gas produced when 70.1 g of iron reacts with hydrochloric acid, at 25 °C and 1 atm, is 30.77 liters.

Step-by-step explanation:

To determine the volume of hydrogen gas produced when 70.1 g of iron reacts completely with hydrochloric acid, we first need to use the balanced chemical equation for the reaction:
Fe(s) + 2HCl(aq) → FeCl2 (aq) + H2(g).

The equation shows that one mole of iron produces one mole of hydrogen gas. Using the molar mass of iron (55.85 g/mol), we can calculate the moles of iron:
70.1 g Fe × (1 mol Fe / 55.85 g Fe) = 1.255 mol Fe.
Since the molar ratio of Fe to H2 is 1:1, 1.255 moles of Fe will produce 1.255 moles of H2.

Given that the reaction is taking place at 25 °C (298 K) and 1 atm, we can use the Ideal Gas Law (PV = nRT) to find the volume. Assuming a value of 0.0821 L
cdot atm/mol
cdot K for the gas constant R, we have:
V = (nRT)/P = (1.255 mol × 0.0821 L
cdot atm/mol
cdot K × 298 K) / 1 atm = 30.77 L of H2.

Answer 2

`\large \boxed{\text{46.1 L}}`

Step-by-step 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

`\text{Moles of Fe } =\text{70.1 g Fe } * \frac{\text{1 mol Fe }}{\text{55.84 g Fe }} =\text{1.255 mol Fe}`

(b) Moles of H₂

The molar ratio is 3 mol H₂:2 mol Fe

`\text{Moles of H$_(2)$}= \text{1.255 mol Fe} * \frac{\text{3 mol H$_(2)$}}{ \text{2 mol Fe}} = \text{1.883 mol H$_(2)$}`

(c) Volume of H₂

We can use the Ideal Gas Law to calculate the volume of hydrogen.

pV = nRT

`\rm V = (nRT)/(p)= \frac{\text{1.883 mol $*$ 0.08206 L$\cdot$atm$\cdot$K$^(-1)$mol$^(-1)*$ 298.15 K}}{\text{ 1 atm}} = \textbf{46.1 L} \\\\\text{You can produce $\large \boxed{\textbf{46.1 L}} $ of H$_(2)$.}`