If a sample of 27.0 g of aluminum metal is added to 333 mL of 3.0 M HCL acid, the volume of hydrogen gas produced at standard temperature and pressure is
a. 2.8 L
b. 5.60 L
c. 11.2 L
d. 22.4 L
6HCl + 2Al>>2AlCl3 + 3H2
moles aluminum=27/27=1mole which means you need 3 moles HCL.
moles HCl=.333*3=1.0 moles
so the limiting reactant is HCL, giventhat, moles H2 generated=1/6 * 3*22.4 liters.
To determine the volume of hydrogen gas produced, we should start by writing a balanced chemical equation for the reaction between aluminum metal and hydrochloric acid:
2 Al + 6 HCl -> 2 AlCl3 + 3 H2
From the balanced equation, we can see that 2 moles of aluminum react with 6 moles of hydrochloric acid to produce 3 moles of hydrogen gas.
Let's calculate the number of moles of aluminum using its molar mass:
Molar mass of aluminum (Al) = 27.0 g/mol
Number of moles of aluminum = mass of aluminum / molar mass of aluminum
= 27.0 g / 27.0 g/mol
= 1 mol
From the balanced equation, we can determine that 1 mole of aluminum reacts to produce 3 moles of hydrogen gas.
Number of moles of hydrogen gas = number of moles of aluminum * (3 moles H2 / 1 mole Al)
= 1 mol * (3 mol H2 / 1 mol Al)
= 3 mol H2
We know that 1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP).
Volume of hydrogen gas = number of moles of hydrogen gas * volume at STP
= 3 mol * 22.4 L/mol
= 67.2 L
Therefore, the volume of hydrogen gas produced at standard temperature and pressure is 67.2 L.
None of the provided answer choices match the calculated value, so none of them are correct.
To answer this question, we need to use the balanced chemical equation for the reaction between aluminum and hydrochloric acid.
The balanced equation is:
2 Al + 6 HCl -> 2 AlCl3 + 3 H2
From the balanced equation, we can see that 2 moles of aluminum react with 6 moles of HCl to produce 3 moles of hydrogen gas.
First, we need to calculate the number of moles of aluminum in the sample. We can do this by dividing the mass of aluminum by its molar mass.
The molar mass of aluminum is 26.98 g/mol. Therefore, the number of moles of aluminum is:
moles of aluminum = 27.0 g / 26.98 g/mol ≈ 1.00 mol
Since the stoichiometry of the balanced equation tells us that 2 moles of aluminum react with 3 moles of hydrogen gas, the number of moles of hydrogen gas produced is:
moles of hydrogen gas = (1.00 mol aluminum) * (3 mol hydrogen gas / 2 mol aluminum) ≈ 1.50 mol
Now, we can apply the Ideal Gas Law to find the volume of gas produced at standard temperature and pressure (STP), which is 0 degrees Celsius (or 273.15 Kelvin) and 1 atmosphere (or 101.3 kilopascals).
The Ideal Gas Law is given by:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L atm / (mol K))
T = temperature (in K)
Since we want to solve for V, we rearrange the equation to:
V = (nRT) / P
Plugging in the values, we get:
V = (1.50 mol) * (0.0821 L atm / (mol K)) * (273.15 K) / (1 atm)
V ≈ 33.60 L
The volume of hydrogen gas produced at STP is approximately 33.60 L.
Since none of the answer choices match exactly, it seems there might be an error in either the question or the calculations. Kinly check your question and calculations again.