$$(a)\\M(NH_3)=17,031\frac{g}{mol}\\n=\frac{m}{M}=\frac{450g}{17,031\frac{g}{mol}}\approx26,422mol\\V_m=\frac{V}{n}\Rightarrow V=V_m\cdot n=22,414\frac{L}{mol}\cdot26,422mol\approx592,223L=0,592223m^3\\(b)\\Q_N=Q\cdot\frac{p\cdot T_N}{p_N\cdot T}\Rightarrow Q=Q_N\cdot\frac{p_N\cdot T}{p\cdot T_N}\\Q_N=4000\frac{m^3}{h}\\p_N=101325Pa\\T=43°C=316,15K\\p=36bar=3,6\cdot10^6Pa\\T_N=0°C=273,15K\\Q=4000\frac{m^3}{h}\cdot\frac{101325Pa\cdot316,15K}{3,6\cdot10^6Pa\cdot273,15K}\approx130,307\frac{m^3}{h}=130307\frac{L}{h}$$