Дано
m(NH4CL) = 107 g
m(Ca(OH)2) = 37 g
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m(NH3)-?
M(NH4CL) = 53.5 g/mol
n(NH4CL) = m/M = 107 / 53.5 = 2 mol
M(Ca(OH)2)= 74 g/mol
n(Ca(OH)2)= m/M= 37/74 = 0.5 mol
n(NH4CL)>n(Ca(OH)2)
37 X
2NH4CL+Ca(OH)2-->CaCL2+2NH3+2H2O M(NH3)= 17 g/mol
74 2*17
X= 37*2*17 / 74 = 17 g
ответ 17 г