Trong 1 mol NH4NO3 có chứa 2 mol N
112kg=112000g
\(n_N=\dfrac{112000}{14}=8000\left(mol\right)\)
\(n_{NH_4NO_3}=\dfrac{1}{2}.n_N=\dfrac{1}{2}.8000=4000\left(mol\right)\)
\(m_{NH_4NO_3}=4000.80=320000\left(g\right)\)
\(m_A=\dfrac{320000.100}{80}=400000\left(g\right)=400\left(kg\right)\)