Question

Consider 2 Al + 6 HCl → 2 AlCl3 + 3 H2 , the reaction of Al with HCl to produce hydrogen gas. What is the pressure of H2 if the hydrogen gas collected occupies 14.0 L at 300.K and was produced upon reaction of 4.50 moles of Al and excess HCl in a process that has a 75.4 percent yield?

Answer 1

The pressure of H2 gas produced in the reaction is 0.717 atm.

Step-by-step explanation:

To determine the pressure of H2 gas, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. In this case, we are given the volume of H2 gas (14.0 L), the number of moles of Al (4.50 moles), and the temperature (300 K). We also know that the reaction has a 75.4 percent yield, so we need to take that into account. First, we need to calculate the number of moles of H2 gas produced in the reaction:

2 moles Al : 3 moles H2 => 4.50 moles Al : x moles H2

x = (4.50 moles Al) * (3 moles H2 / 2 moles Al) = 6.75 moles H2

Next, we can use the ideal gas law equation to calculate the pressure:

PV = nRT

P * 14.0 L = (6.75 moles)(0.0821 L atm/mol K)(300 K)

P = (6.75 moles * 0.0821 L atm/mol K * 300 K) / 14.0 L = 0.717 atm

Therefore, the pressure of H2 gas is 0.717 atm.