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ASIKE002
@ASIKE002
July 2022
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при взаимодействие хлора и водорода образовалось 10,08л хлороводорода. Вычислите массы и количества веществ, вступивших в реакцию.
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eozin
H2 + Cl2 = 2HCl
n=? n=? n=
Vm=22.4 l/mol
m=? m=? V=10.8 l
n(HCl)=V/Vm = 10.8/22.4 = 0.482mol
n(H2) = n(HCl)/2 = 0.482/2 = 0.241mol
n(Cl2) = n(HCl)/2 = 0.482/2 = 0.241mol
m(H2) = n(H2)*M(H2)= 0.241*2 = 0.482g
m(Cl2)= n(Cl2)*M(Cl2)=0.241*71= 17.11g
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Answers & Comments
n=? n=? n=
Vm=22.4 l/mol
m=? m=? V=10.8 l
n(HCl)=V/Vm = 10.8/22.4 = 0.482mol
n(H2) = n(HCl)/2 = 0.482/2 = 0.241mol
n(Cl2) = n(HCl)/2 = 0.482/2 = 0.241mol
m(H2) = n(H2)*M(H2)= 0.241*2 = 0.482g
m(Cl2)= n(Cl2)*M(Cl2)=0.241*71= 17.11g