Respuesta :
Answer:
1.24 L of H₂ at STP .
Explanation:
2Al(s) +6HCl(aq) → 2AlCl₃(aq) + 3H₂(g)
2 moles 3 x 22.4 L
2 x 27 g of Al reacts to give 3 x 22.4 L of H₂ at STP .
1 g of Al will react to give 3 x 22.4 / ( 2 x 27 ) L of H₂ at STP .
= 1.24 L of H₂ at STP .
The volume of hydrogen produced by 1 grams of Al has been 1.24 L.
The balanced chemical reaction has been given as:
[tex]\rm 2\;Al\;+\;6\;HCl\;\rightarrow\;2\;AlCl_3\;+\;3\;H_2[/tex]
From the equation, 2 moles of Aluminum gives 3 moles of Hydrogen
The mass of the compound from moles can be given as:
Mass = moles × molecular mass
Mass of 2 moles Al = 2 × 27 g
Mass of 2 moles Al = 54 g
Mass of 3 moles hydrogen = 3 × 2 g
Mass of 3 moles hydrogen = 6 g
From the equation,
54 g aluminum gives = 6 grams hydrogen
[tex]\rm 1\;gram\;aluminum\;=\;\dfrac{6}{54}\;\times\;1[/tex]
1 gram Aluminum = 0.11 grams hydrogen
The mass of hydrogen produced by 1 gram Al has been 0.11g. The moles equivalent to 0.11g hydrogen has been given as:
Mass = moles × molecular mass
0.11 g = moles × 2 g/mol
Mole of hydrogen = 0.055 mol
The moles of hydrogen produced by 1 gram of Al has been 0.055 mol.
According to the ideal gas equation, any gas at STP has 1 mole equivalent to 22.4 L. So,
1 mol = 22.4 L
0.055 mol = 0.055 × 22.4 L
0.055 mol = 1.244 L.
The volume of hydrogen produced by 1 grams of Al has been 1.24 L.
For more information about volume at STP, refer to the link:
https://brainly.com/question/11676583
