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TalismDascha
@TalismDascha
June 2022
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СРОЧНО. ДАЮ 65 БАЛЛОВ. РЕШИТЕ, ПОЖАЛУЙСТА ЗАДАЧИ ПО ХИМИИ
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Alexei78
1.
дано
m(NH4CL) = 20 g
m(Ca(OH)2) = 20 g
η = 98%
-------------------------
m(практNH3)-?
M(MH4CL)= 53.5 g/mol
n(NH4CL) = m/M= 20/53.5 = 0.37 mol
M(Ca(OH)2)= 74 g/mol
n(Ca(OH)2)= m/M= 20/74 = 0.27 mol
n(NH4CL)>n(Ca(OH)2)
M(NH3)= 17 g/mol
20 X
2NH4CL+Ca(OH)2-->CaCL2+2NH3↑+2H2O
74 2*17
X = 20*2*17 / 74 = 9.2 g
m(практ NH3) = 9.2*98% / 100% = 9 g
ответ 9 г
2.
дано
V(N2) = 56 L
+H2
η(NH3)= 50%
-------------------
Vпракт(NH3)-?
m(NH3)-?
56 X
N2+3H2-->2NH3
1mol 2mol
X = 56*2 / 1 = 112 L
Vпракт(NH3) = 112*50% / 100% = 56 L
n(NH3) = V(практ NH3)/Vm = 56/22.4 = 2.5 mol
m(NH3) =n(NH3)*M(NH3) = 2.5*17 = 42.5 g
ответ 56 л, 42.5 г
3.
дано
m(NH4CL) = 107 g
V(NH3)= 38 L
-----------------------
φ(NH3)-?
107 X
NH4CL-->NH3+HCL M(NH4CL) = 53.5 g/mol Vm = 22.4 L/mol
53.5 22.4
X = 107*22.4 / 53.5 = 44.8 L
φ(NH3) = V(прак) / V(теор) * 100% = 38 / 44.8 * 100% = 84.8%
ответ 84.8%
2 votes
Thanks 1
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Answers & Comments
дано
m(NH4CL) = 20 g
m(Ca(OH)2) = 20 g
η = 98%
-------------------------
m(практNH3)-?
M(MH4CL)= 53.5 g/mol
n(NH4CL) = m/M= 20/53.5 = 0.37 mol
M(Ca(OH)2)= 74 g/mol
n(Ca(OH)2)= m/M= 20/74 = 0.27 mol
n(NH4CL)>n(Ca(OH)2)
M(NH3)= 17 g/mol
20 X
2NH4CL+Ca(OH)2-->CaCL2+2NH3↑+2H2O
74 2*17
X = 20*2*17 / 74 = 9.2 g
m(практ NH3) = 9.2*98% / 100% = 9 g
ответ 9 г
2.
дано
V(N2) = 56 L
+H2
η(NH3)= 50%
-------------------
Vпракт(NH3)-?
m(NH3)-?
56 X
N2+3H2-->2NH3
1mol 2mol
X = 56*2 / 1 = 112 L
Vпракт(NH3) = 112*50% / 100% = 56 L
n(NH3) = V(практ NH3)/Vm = 56/22.4 = 2.5 mol
m(NH3) =n(NH3)*M(NH3) = 2.5*17 = 42.5 g
ответ 56 л, 42.5 г
3.
дано
m(NH4CL) = 107 g
V(NH3)= 38 L
-----------------------
φ(NH3)-?
107 X
NH4CL-->NH3+HCL M(NH4CL) = 53.5 g/mol Vm = 22.4 L/mol
53.5 22.4
X = 107*22.4 / 53.5 = 44.8 L
φ(NH3) = V(прак) / V(теор) * 100% = 38 / 44.8 * 100% = 84.8%
ответ 84.8%