When 35 mL of 0.92 M sulfuric acid reacts with excess Al, how many L of hydrogen gas is formed at 23 °C and a pressure of 745 mm Hg?
2Al + 3H2SO4 ----> Al2(SO4)3 + 3H2

Answers

Answer 1

When 35 mL of 0.92 M sulfuric acid reacts with excess aluminum, approximately 0.823 L of hydrogen gas is formed at 23 °C and a pressure of 745 mm Hg.

To determine the volume of hydrogen gas formed, we need to use the ideal gas law equation:

PV = nRT

Where:

P = pressure of the gas (in atm)

V = volume of the gas (in liters)

n = number of moles of gas

R = ideal gas constant (0.0821 L·atm/(mol·K))

T = temperature (in Kelvin)

Convert the given pressure to atm:

745 mm Hg * (1 atm / 760 mm Hg) = 0.9797 atm

Convert the given volume to liters:

35 mL * (1 L / 1000 mL) = 0.035 L

Calculate the number of moles of hydrogen gas produced.

From the balanced equation:

2 Al + 3 H₂SO₄ → Al₂(SO₄)₃ + 3 H₂

We can see that 3 moles of hydrogen gas are produced for every 3 moles of H₂SO₄.

Given that the concentration of sulfuric acid is 0.92 M and the volume used is 35 mL, we can calculate the number of moles of H₂SO₄ used:

moles of H₂SO₄ = concentration * volume

moles of H₂SO₄ = 0.92 M * 0.035 L = 0.0322 moles

Therefore, the number of moles of hydrogen gas produced is also 0.0322 moles.

Then convert the temperature to Kelvin:

23 °C + 273.15 = 296.15 K

Plug the values into the ideal gas law equation to find the volume of hydrogen gas:

PV = nRT

(0.9797 atm) * V = (0.0322 mol) * (0.0821 L·atm/(mol·K)) * (296.15 K)

Solving for V:

V = (0.0322 mol) * (0.0821 L·atm/(mol·K)) * (296.15 K) / (0.9797 atm)

V ≈ 0.823 L

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Answers

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