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krisvayt
@krisvayt
August 2021
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73г 10%-го раствора соляной кислоты взаимодействуют с избытком оксида магния. Какова масса образовавшейся соли?
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vipmaster2011
MgO+2HCI=MgCI2+H2O
m(ве-ва)(HCI)=73*0,1=7,3гр
M(HCI)=36,5гр
n(HCI)=m/M=7,3/36,5=0,2моль
M(MgCI2)=95г/моль
m(MgCI2)=m*M=95*0,2=19гр
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Answers & Comments
m(ве-ва)(HCI)=73*0,1=7,3гр
M(HCI)=36,5гр
n(HCI)=m/M=7,3/36,5=0,2моль
M(MgCI2)=95г/моль
m(MgCI2)=m*M=95*0,2=19гр