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Пропускаючи карбон(IV) оксид крізь 28,5 г 12%-го розчину барій гідроксиду добули 3,743 г. барій карбонату. Обчисліть масову частку виходу барій карбонату у % від теоретично можливого ?
Пропускаючи амоніак крізь 31,5 г 20% го розчину нітратої кислоти добули 6,4 г. амоній нітрату. Яка масова частка виходу амоній нітрату у % від теоретично можливого ?

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Объяснение:

K2+2H2O=2KOH + H2

N2O5+10HCl=2NCl5 + 5H2O

2Ca+H2SO4=Ca2SO4 + H2

KOH+HNO2=KNO2 + H2O

3B2O3+6Fe(OH)3= 2Fe3(BO3)3 + 9H2O

Zn+2H2O=Zn(OH)2 + H2

2Fe(OH)3=Fe2O3+H2O

BaO+2HNO3=BA(NO3)2 + H2O

2NaOH+CuSO4=Na2SO4 + Cu(OH)2

2Na+2H2O=2NaOH + H2

Mn2O7+7Mg=2Mn + 7MgO

P205+2H2O=2H3PO4

Cu(OH)2+2NaOH=CuO2 + 2H2O + 2Na

Ba(OH)2+2NaCl=BaCl2 + 2NaOH

2Au+2H2O=2AuOH + H2

CaCO3=CaO + CO2

Fe(OH)3+3HClO4=Fe(ClO4)3 + 3H2O

AgNO3+2NaCl=Na2NO3 + AgCl2

Cu+2HCl=CuCl2 + H2

Ba(OH)2+NaCl уже выше такая реакция была

ZnO+SO2=ZnSO3

2CH4+3O2=2CO + 4H2O

2CuO+2H2O=2CuOH2 + O2

Fe+2CuCl=2Cu + FeCl2

Fe2O3+Na2O=2NaFeO2

BaCl2+NaSO4=BaSO4 + 2NaCl

Na2SO4+2KCl=2NaCl + K2SO4

3H2SO4+2HNO3=4H2O + 2NO + 3SO4

Na2SO4+CuCl2=2NaCl + CuSO4

CO2+2NaOH=Na2CO3 + H2O