What volume of H2 (g) (at 750. mm Hg and 25 C) is produced from 50.0 mL of 0.214 M H2SO4 and 0.300 g of Al?
To find the volume of H2 gas produced, we can use the balanced chemical equation for the reaction between aluminum (Al) and sulfuric acid (H2SO4):
2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2.
From the balanced equation, we can see that 2 moles of Al react with 3 moles of H2SO4 to produce 3 moles of H2 gas. Based on the given molar concentration of H2SO4 (0.214 M) and the volume of H2SO4 (50.0 mL), we can determine the amount of moles of H2SO4 used.
First, convert the volume of H2SO4 from milliliters (mL) to liters (L):
50.0 mL = 50.0 mL * (1 L / 1000 mL) = 0.050 L.
Next, use the molar concentration (0.214 M) to calculate the number of moles of H2SO4:
moles of H2SO4 = molar concentration * volume of H2SO4
= 0.214 M * 0.050 L
= 0.0107 moles.
Since the stoichiometry of the balanced equation shows that 2 moles of Al react with 3 moles of H2, we can determine the number of moles of H2 produced:
moles of H2 = (moles of H2SO4 * 3) / 2
= (0.0107 moles * 3) / 2
= 0.01605 moles.
Now, let's apply the ideal gas law to calculate the volume of H2 gas produced. The ideal gas law is given by:
PV = nRT,
where P represents the pressure (750. mm Hg), V is the volume of H2 gas we want to find, n is the number of moles of H2 (0.01605 moles), R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature (25 °C or 298 K).
Rearrange the equation to solve for V:
V = (nRT) / P
= (0.01605 moles * 0.0821 L·atm/mol·K * 298 K) / 750. mm Hg.
Next, convert the pressure from millimeters of mercury (mm Hg) to atmospheres (atm):
750. mm Hg = 750. mm Hg * (1 atm / 760 mm Hg) = 0.987 atm.
Substitute the values into the equation:
V = (0.01605 moles * 0.0821 L·atm/mol·K * 298 K) / 0.987 atm
≈ 0.395 L.
Therefore, the volume of H2 gas produced is approximately 0.395 liters.