CTTQ:$N_{x}H_{y}Cl_{z}$
Ta có:$\frac{x.14}{26,17}=\frac{y.1}{2,14}=\frac{35,5.z}{71,69}=\frac{M_{N_{x}H_{y}Cl_{z}}}{100}\\\Rightarrow \frac{14.x}{26,17}=\frac{y}{2,14}=\frac{35,5.z}{71,69}=\frac{53,5}{100}=0,535\\$
$\Rightarrow$x=$\frac{0,535.26,17}{14}\approx 1$
y=$0,535.2,14\approx 1$
z=$\frac{0,535.71,69}{35,5}\approx 1$
Vậy CTHH của hợp chất A là:NHCl