2AL+6H-->2AL+3H.
What volume of hydrogen gas at STP would be evolved in a reaction of 36.0mg of 98% pure aluminum with the hydrogen being an ideal?
I got the ml of hydrogen, but I don't know how to work it with AL being 98%pure.
equation does not make sense as typed.
Perhaps:
2Al + 6 H+ ---> 2 Al+3 + 3 H2(gas)
The aluminum is 98 % pure
so we really have .98*36 =35.3 mg of Al
How many moles of Al is that?
Al is 27 grams/mol
so we have 35.3 *10^-3 g / 27 g/mol = 1.31 * 10^-3 mol of Al
I get 3 moles of H2 for every 2 moles of Al
so moles of H2 = (3/2)(1.31*10^-3) = 1.96 *10^-3 mol of H2
at STP an ideal gas is about 22.4 liters/ mole
so
1.96 *10^-3 mol * 22.4 L/mol = 43.9^10^-3 L = .0439 L = 43.9 mL
Thanks so much! Sorry about the equation, it was 3 H2.
To find the volume of hydrogen gas evolved in the reaction, we need to perform a stoichiometric calculation. Here's how you can do it:
1) First, calculate the moles of aluminum (Al) present in the reaction using the given mass and the molar mass of aluminum (26.98 g/mol):
Moles of Al = Mass of Al / molar mass of Al
Moles of Al = 36.0 mg / (0.98 * 26.98 g/mol) (since aluminum is 98% pure)
Moles of Al = 1.3309 mmol
2) Then, determine the stoichiometric ratio between aluminum and hydrogen in the balanced chemical equation. From the equation:
2Al + 6H₂ → 2AlH₃
It is clear that 6 moles of hydrogen gas (H₂) react with 2 moles of aluminum (Al).
3) Using the stoichiometric ratio, calculate the moles of hydrogen produced:
Moles of H₂ = (Moles of Al / 2) * 6
Moles of H₂ = (1.3309 mmol / 2) * 6
Moles of H₂ = 3.9927 mmol
4) Convert the moles of hydrogen gas to volume at STP (Standard Temperature and Pressure). At STP, the molar volume of an ideal gas is 22.4 L/mol.
Volume of H₂ = Moles of H₂ * Molar volume at STP
Volume of H₂ = 3.9927 mmol * (22.4 L/mol)
Volume of H₂ = 89.36 ml
So, the volume of hydrogen gas evolved in the reaction with 36.0 mg of 98% pure aluminum would be approximately 89.36 ml at STP.