Question

Answer 1

66. Số mol của 5 tấn HNO3 63% là : \(\dfrac{5.63\%}{63}=0,05\left(mol\right)\)

Vì H=70% => \(n_{NO_2}=\dfrac{0,05}{70\%}=\dfrac{1}{14}\left(mol\right)\) 

Vì H=50% => \(n_{NO}=\dfrac{1}{14}:50\%=\dfrac{1}{7}\left(mol\right)\) 

Vì H=23% => \(n_{NH_3}=\dfrac{1}{7}:23\%=\dfrac{100}{161}\left(mol\right)\) 

=>\(V_{NH_3}=\dfrac{100}{161}.22,4=13,913\left(m^3\right)\)

Answer 2

\(H\%=23\%+50\%+70\%=143\%\)

\(m_{HNO_3}=5\cdot10^3\cdot63\%=3150\left(kg\right)\)

\(n_{HNO_3}=\dfrac{3150}{63}=50\left(kmol\right)\)

Bảo toàn nguyên tố N : 

\(n_{NH_3}=n_{HNO_3}=50\left(mol\right)\)

\(\Rightarrow n_{NH_3\left(bđ\right)}=50\cdot143\%=71.5\left(kmol\right)\)

\(V_{NH_3}=71.5\cdot10^3\cdot22.4=1601.6\left(l\right)=1.6016\left(m^3\right)\)