[tex]m=270\:g\:(Al)[/tex]
p = 90% (Al)
ρ = 1,104 g/mL (H₂SO₄)
c = 20% (H₂SO₄)
__________________
V = ? (H₂SO₄)
Reacția chimică:
2Al + 3H₂SO₄ → Al(SO₄)₃ + 3H₂↑
[tex]m_{Al} =m_{impur}=270\:g\\[/tex]
[tex]\bf\red{\boxed{m_{pur}=\frac{p\times m_{impur}}{100}}}[/tex]
[tex]m_{pur}=\frac{90\times270}{100}=243\:g[/tex]
[tex]M_{H_2SO_4}=2A_H+A_S+4A_O=2+32+4\times16=98\:g/mol\\[/tex]
2 × 27 g/mol ............ 3 × 98 g/mol
243g ............ x g
[tex]\implies x=\frac{243\times3\times98}{2\times27}=1323\:g\:H_2SO_4\\\[/tex]
[tex]m_s=\frac{m_d\times100}{c}=\frac{1323\times100}{20}=6615\:g\\[/tex]
[tex]V=\frac{m}{\rho}=\frac{6615}{1,104}=\bf5991,8\:mL\\[/tex]
Succes! ❀
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