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ASIKE002
@ASIKE002
July 2022
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При взаимодействии хлора и водорода образовалось 10,08л хлороводорода.Вычислите массы и количества веществ, вступивших в реакцию
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Alexei78
V(HCL)=10.08 L
--------------------
m(H2)-?
m(CL2)-?
n(H2)-?
n(CL2)-?
X Y 10.08
H2+CL2-->2HCL M(H2)=2 g/mol M(CL2)=71 g/mol Vm=22.4L/mol
2 71 2*22.4
X(H2)=2*10.08 / 44.8 = 0.45 g
Y(CL2)= 71*10.08 / 44.8 = 16 g
n(h2)=m/M=0.45 / 2 = 0.225 mol
n(CL2)=m/M=16/71 = 0.225 mol
ответ n(H2)=n(CL2) = 0.225 mol, m(H2)=0.45 g , m(CL2)=16 g
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Answers & Comments
--------------------
m(H2)-?
m(CL2)-?
n(H2)-?
n(CL2)-?
X Y 10.08
H2+CL2-->2HCL M(H2)=2 g/mol M(CL2)=71 g/mol Vm=22.4L/mol
2 71 2*22.4
X(H2)=2*10.08 / 44.8 = 0.45 g
Y(CL2)= 71*10.08 / 44.8 = 16 g
n(h2)=m/M=0.45 / 2 = 0.225 mol
n(CL2)=m/M=16/71 = 0.225 mol
ответ n(H2)=n(CL2) = 0.225 mol, m(H2)=0.45 g , m(CL2)=16 g