50 g 50g XL 
2NH4CL+Ca(OH)2->CaCL2+2NH3+2H2O
2*53.5 74 2*22.4
M(NH4CL)=53.5 g/mol
M(Ca(OH)2=74 g/mol
Vm=22.4L/mol

n(NH4CL)=50/53.5=0.93 mol
n(Ca(OH)2)=50/74=0.68 mol
n(NH4CL)>n(Ca(OH)2)
50/74 =X/44.8
X=30.27 L
ответ 30.27 л