Ответ:
дано
m(ppa HCL) = 365 g
W(HCL) = 10%
+Al
-------------------------
V(H2)-?
m(HCL) = 365 * 10% / 100% = 36.5 g
2Al+6HCL-->2AlCL3+3H2
M(HCL) = 36.5 g/mol
n(HCL) = m/M = 36.5 / 36.5 = 1 mol
6n(HCL) = 3n(H2)
n(H2) = 3*1 / 6 = 0.5 mol
V(H2) = n*Vm =0.5 * 22.4 = 11.2 L
ответ 11.2 л
Объяснение:
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Answers & Comments
Ответ:
дано
m(ppa HCL) = 365 g
W(HCL) = 10%
+Al
-------------------------
V(H2)-?
m(HCL) = 365 * 10% / 100% = 36.5 g
2Al+6HCL-->2AlCL3+3H2
M(HCL) = 36.5 g/mol
n(HCL) = m/M = 36.5 / 36.5 = 1 mol
6n(HCL) = 3n(H2)
n(H2) = 3*1 / 6 = 0.5 mol
V(H2) = n*Vm =0.5 * 22.4 = 11.2 L
ответ 11.2 л
Объяснение: