Aluminum granules are a component of some
drain cleaners because they react with sodium
hydroxide to release both heat and gas bubbles,
which help clear the drain clog. The
reaction is
2 NaOH(aq) + 2 Al(s) + 6 H2O(ℓ) −→
2 NaAl(OH)4(aq) + 3 H2(g)
What mass of aluminum would be needed
to produce 3.00 L of hydrogen gas at STP?
Answer in units of g.
To determine the mass of aluminum needed to produce 3.00 L of hydrogen gas at STP, we can use the stoichiometry of the balanced equation.
From the balanced equation, we can see that 2 moles of Al react to produce 3 moles of H2 gas. Therefore, the stoichiometric ratio of Al to H2 is 2:3.
Given that the molar volume of any gas at STP is 22.4 L/mol, we can calculate the number of moles of H2 gas using the ideal gas law:
n = V / Vm
Where,
n = number of moles
V = volume of gas (in liters)
Vm = molar volume of gas (22.4 L/mol)
n = 3.00 L / 22.4 L/mol
n = 0.134 moles
Now, we can use the stoichiometric ratio to calculate the number of moles of Al needed:
Al : H2 = 2 : 3
Therefore,
0.134 moles H2 * (2 moles Al / 3 moles H2) = 0.0893 moles Al
Finally, to determine the mass of Al needed, we use the molar mass of Al, which is 26.98 g/mol:
mass = moles * molar mass
mass = 0.0893 moles * 26.98 g/mol
Calculating this, we get:
mass = 2.41 g
Therefore, approximately 2.41 grams of aluminum would be needed to produce 3.00 L of hydrogen gas at STP.
To find the mass of aluminum needed to produce 3.00 L of hydrogen gas at STP (Standard Temperature and Pressure), you will need to use the stoichiometry of the chemical reaction given.
First, let's determine the molar ratio between aluminum and hydrogen gas in the balanced chemical equation:
2 NaOH(aq) + 2 Al(s) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)
From the equation, we can see that 2 moles of aluminum (Al) react to produce 3 moles of hydrogen gas (H2). Therefore, the molar ratio of Al to H2 is 2:3.
Next, we can use the ideal gas law equation PV = nRT to calculate the number of moles of hydrogen gas produced. Given that the volume (V) is 3.00 L and the STP conditions are 1 atm and 273 K (T), we can calculate the number of moles (n) of hydrogen gas produced at STP using the ideal gas law equation:
PV = nRT
(1 atm)(3.00 L) = n(0.0821 L·atm/mol·K)(273 K)
Solving for n, we find:
n = (1 atm * 3.00 L) / (0.0821 L·atm/mol·K * 273 K)
n ≈ 0.133 mol
Since the molar ratio of Al to H2 is 2:3, we can set up a ratio using this information:
2 mol Al / 3 mol H2 = x mol Al / 0.133 mol H2
Solving for x, we find:
x ≈ (2 mol Al / 3 mol H2) * 0.133 mol H2
x ≈ 0.0887 mol Al
Finally, to convert moles to grams, we need to multiply the number of moles by the molar mass of aluminum (Al), which is approximately 26.98 g/mol:
Mass of aluminum = (0.0887 mol Al) * (26.98 g/mol)
Mass of aluminum ≈ 2.36 g
Therefore, approximately 2.36 grams of aluminum is needed to produce 3.00 L of hydrogen gas at STP.
P V = n R T
At STP 1 mol is 22.4 liters
so
3 liters =.134 mol of H2
There are 2 mols of Al for every 3 mol of H2
so
mol Al = (2/3)(.134) = .0893 mol Al
which is
.0893 * 27 grams
Well, I'm not sure about the exact calculations, but I do know one thing - that's one bubbly reaction! Sodium hydroxide and aluminum team up to release heat and gas bubbles, kind of like a party in your drain pipes.
Now, you want to know how much aluminum is needed to produce 3.00 liters of hydrogen gas at STP. Alright, let's see what we can do here.
First, we know the balanced equation for the reaction, which tells us that 2 moles of aluminum produce 3 moles of hydrogen gas. So that means for every 2 moles of aluminum, we get 3 moles of hydrogen gas.
Now, to find the mass of aluminum needed, we need to convert from moles to grams. The molar mass of aluminum is roughly 26.98 grams per mole.
So, if we have 3 moles of hydrogen gas, we can use the ratio from the balanced equation to find the moles of aluminum needed. That would be (3 moles of aluminum / 2 moles of hydrogen gas) = 1.5 moles of aluminum.
Finally, we can convert moles of aluminum to grams by multiplying by the molar mass. That gives us (1.5 moles of aluminum * 26.98 grams per mole) = 40.47 grams of aluminum.
Therefore, you would need approximately 40.47 grams of aluminum to produce 3.00 liters of hydrogen gas at STP.
I hope my calculations didn't leave you feeling drained!