Подобная задача
При взаимодействии 8 г оксида серы VI с избытком раствора гидроксида калия получили 174 г раствора средней соли. Вычислите массовую долю соли в полученном растворе
8г---------------------------------174г
SO3 + 2KOH = К2SO4 + H2O
v=1моль--------------------v=1моль
М=80г/моль--------------М=174г/моль
m=80г-----------------------m=174
найдем массу соли. Для этого сост. пропорцию
8г-----хг
80г---174г
х=(174г*8г) /80г=17,4г
теперь найдем масссовую долю соли. Нам известна масса раствора=174г и масса соли=17,4г
w%(соли) =(17,4г/174г) *100%=10%
ответ: 10%
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\4Mg 4Mg + 5H2SO4 -----> 4MgSO4 + H2S + 4H2O
Mg(0) -2e ---> Mg(+2) _____________4 востановитель
S(+6) + 8e ---> S(-2)_______________1окислитель
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C → CO2 → NaHCO3 → Na2CO3 → CO2 → CO
C + H2O(пар) = CO2 + H2
CO2 + NaOH(разб. ) = NaHCO3
NaHCO3 + NaOH(конц. ) = Na2CO3 + H2O
Na2CO3 = Na2O + CO2
CO2 + C = 2CO
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NH3→ (NH4)3PO4 →NH3 → NO →NO2→HNO3
1.(NH4)2HPO4 + NH3•H2O(конц. ) = (NH4)3PO4 + H2O
2. (NH4)3PO4 = NH3 + (NH4)2HPO4
3.4NH3 + 5O2 = 4NO + 6H2O
4. NO + O3 = NO2 + O
5. H2 + 2NO2 = HNO2 + HNO3
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