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Zinc reacts with aqueous sulfuric acid to form hydrogen gas:

Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
In an experiment, 201 mL of wat Hy is collected over water at 27°C and a barometric pressure of 733 torr. The vapor pressure of water at 27°C is 26.74 torr. How many grams of H₂ are produced?
a. 20.0 g
b. 1.80 x 10⁻² g
c. 130 g
d. 8.61 x 10⁻³ g
e. 1.52 x 10⁻² g

1 Answer

5 votes

Final answer:

To solve this problem, we must first determine the hydrogen gas's pressure, then use the Ideal Gas Law to find the number of moles of H₂, and finally calculate the mass of hydrogen produced, which is approximately 1.52 x 10⁻² grams.

Step-by-step explanation:

To determine the amount of hydrogen gas produced, we need to use the combined gas law and the concept of stoichiometry. First, we find the pressure of hydrogen gas alone by subtracting the water vapor pressure from the total barometric pressure (733 torr - 26.74 torr = 706.26 torr).

We must then convert the pressure to atmospheres, where 1 atm = 760 torr, resulting in an H₂ pressure of approximately 0.9298 atm.

We use the Ideal Gas Law, PV = nRT, where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles of gas, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin (27°C + 273.15 = 300.15 K). By substituting the known values, we can solve for the number of moles of H₂.

n = (P×V) / (R×T)
n = (0.9298 atm × 0.201 L) / (0.0821 L·atm/(mol·K) × 300.15 K)
n = 0.00768 moles of H₂

Now, knowing that the molar mass of H₂ is approximately 2.016 g/mol, we can find the mass in grams:

mass = n × molar mass
mass = 0.00768 moles × 2.016 g/mol
mass = 0.0155 g of H₂

Rounding to three significant figures, the mass is approximately 0.0155 g, which is closest to option e, 1.52 x 10⁻² g.

User Moaz Khan
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