5,4 g de aluminiu reactioneaza cu cantitatea necesara de sulf. Calculati masa de sulf necesara si masa de sulfura de aluminui necesara.​

Bună!

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[tex] \\ 2Al + 3S→Al_2S_3 \\ \\ M_{Al_2S_3} = 2 \times 27 + 3 \times 32 = 150 \: g/mol \\ \\ 2 \times 27g \: Al \: .............. \: 3 \times 32g \: S \\ 5.4 g \: Al \: ................ \: x \: g \: S \\ x = \frac{3 \times 32 \times 5.4}{2 \times 27} = \frac{518.4}{54} = \blue{9.6g \: S} \\ \\ 2 \times 27g \: Al ............... 150g \: Al_2S_3 \\ 5.4 \: g \: Al \: ................. \: y \: g \: Al_2S_3 \\ y = \frac{150 \times 5.4}{2 \times 27} = \purple{ 15g \: Al_2S_3}[/tex]

Răspuns:

m = 19,2g S

m = 30g Al₂S₃

Rezolvare:

m = 5,4g (Al)

m = ? (S)

m = ? (Al₂S₃)

Scriem ecuația reacției chimice:

2Al + 3S → Al₂S₃

Scriem masele molare:

[tex]M_{Al}=27\:g/mol\\M_S=32\:g/mol\\M_{Al_2S_3}=2A_{Al}+3A_S=2\times27+3\times32=54+96=150\:g/mol\\[/tex]

2 × 27g Al ............ 3 × 32g S

2 × 5,4g Al ............ x g S

54g Al ........... 96g S

10,8g Al .......... x g S

[tex]x=\frac{96\times10,8}{54}=\frac{1036,8}{54}=\bf19,2g\:S\\[/tex]

54g Al .......... 150g Al₂S₃

10,8g Al ......... x g Al₂S₃

[tex]y=\frac{10,8\times150}{54}=\frac{1620}{54}=\bf30g\:Al_2S_3\\[/tex]

Succes! ❀

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