1.\(n_{NH_3}=\dfrac{m_{NH_3}}{M_{NH_3}}=\dfrac{3,4}{17}=0,2\left(mol\right)\)
\(V_{NH_3}=n_{NH_3}.22,4=0,2.22,4=4,48\left(l\right)\)
\(n_{SO_4}=\dfrac{m_{SO_4}}{M_{SO_4}}=\dfrac{9,6}{96}=0,1\left(mol\right)\)
\(V_{SO_4}=n_{SO_4}.22,4=0,1.22,4=2,24\left(l\right)\)
\(V_{hh}=V_{NH_3}+V_{SO_4}=4,48+2,24=6,72\left(l\right)\)