Verified answer

Ответ:

Al4C3+12H2O-->4Al(OH)3+3CH4

2CH4-->C2H2+3H2

C2H2+H2-->C2H4

C2H4+HCL-->C2H5CL

2C2H5CL+2Na-->C4H10+2NaCL

C4H10-->C4H8+H2

дано

m(С7H16) = 50 g

η(H2) = 50%

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V пр.(H2) - ?

C7H16 --> C7H8 + 4H2

M(C7H16) =100 g/mol

n(C7H16) = m/M = 50 / 100 = 0.5 mol

n(C7H16) = 4n(H2)

n(H2) = 4*0.5 = 2 mol

V(H2) = n(H2) * Vm = 2 * 22.4= 44.8 L - теоретический

V пр(H2) = 44.8 *50% / 100% = 22.4 L

ответ 22.4 л

Объяснение: