contestada

What mass of aluminum (in g) would be
required to completely react with 1.30 L of
0.350 M HCl in the following chemical
reaction?
2 Al(s) + 6 HCl(aq) → 2 AICI: (aq) + 3 H2(g)

Respuesta :

ardni313

Mass of Aluminum (in g) : 4.104

Further explanation

Reaction

2 Al(s) + 6 HCl(aq) → 2 AICI₃ (aq) + 3 H₂(g)

mol HCl :

[tex]\tt 1.3\times 0.35=0.455[/tex]

mol Al :

[tex]\tt \dfrac{2}{6}\times 0.455=0.152[/tex]

mass of Aluminum :

[tex]\tt 0.152\times 27~g/mol=4.104~g[/tex]

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