Предмет: Химия, автор: krso

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Вычислите объем (н.у.) аммиака, который можно получить при взаимодействии сульфата аммония масой 33 г и гидроксида кальция 37 г.

Ответы

Автор ответа: Аноним
0
Дано:
m((NH4)2SO4)=33г
m(Ca(OH)2) = 37г
Найти:
V(NH3)
Решение:
(NH4)2SO4 + Ca(OH)2 = 2NH3↑ + 2H2O + CaSO4
n((NH4)2SO4)= m/M = 33/132=0,25 моль
n(Ca(OH)2) = m/M = 37/74 = 0,5 моль
сульфат аммония в недостатке
n(NH3) = 0,5 моль
V(NH3) = Vm*n= 22,4*0,5= 11,2 л


Ответ: V(NH3) = 11,2 л



луйчший не забудь:3
Автор ответа: lyubavasil
0
(NH4)2SO4+Ca(OH)2=2 NH3+2H2O+CaSO4 задача на избыток и недостаток
n=m/M(Mr)   M=Mr   Mr(NH4)2SO4=132    M=132г/моль
n(NH4)2SO4= 33/132=0,25 моль
n(Ca(OH)2=37/74=0,5 моль     MrCa(OH)2=74     M=74г/моль
 Составляем пропорцию, для определения недостатка, за х принимаем количество гидроксида кальция
0,25/1=x/1     x=0,25(  а у нас 0,5 моль) , следовательно Ca(OH)2    в избытке , решаем по (NH4)SO4
n(NH3) =0,25*2=0,5  моль, по условию уравнения , соотношение 1:2
V= n*Vm        V(NH3) = 0,5*22,4=11,2л
Ответ:11,2л
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