Ответ:
дано
m(Al) = 5.4 g
-----------------------
V(H2) - ?
2Al+6HCL-->2AlCL3+3H2
M(AL) = 27 g/mol
n(Al) = m/M = 5,4 / 27 = 0.2 mol
2n(Al) = 3n(H2)
n(H2) = 3*0,2 / 2 = 0.3 mol
V(H2) = n*Vm = 0.3 * 22.4 = 6.72 L
ответ 6.72 л
Объяснение:
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Answers & Comments
Ответ:
дано
m(Al) = 5.4 g
-----------------------
V(H2) - ?
2Al+6HCL-->2AlCL3+3H2
M(AL) = 27 g/mol
n(Al) = m/M = 5,4 / 27 = 0.2 mol
2n(Al) = 3n(H2)
n(H2) = 3*0,2 / 2 = 0.3 mol
V(H2) = n*Vm = 0.3 * 22.4 = 6.72 L
ответ 6.72 л
Объяснение: