\(a.n_{CO_2}=\dfrac{m_{CO_2}}{M_{CO_2}}=\dfrac{4,4}{44}=0,1\left(mol\right)\\ \Rightarrow V_{CO_2}=n_{CO_2}.22,4=0,1.22,4=2,24\left(l\right)\\ n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{3,2}{32}=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2}=n_{O_2}.22,4=0,1.22,4=2,24\left(l\right)\)
\(n_{H_2}\) = 2mol
=>\(V_{H_2}\) = 2. 22,4 = 44,8l
\(n_{O_2}\) = 0,0875 mol
=>\(V_{O_2}\) = 0,0875 . 22,4 = 1,96l
\(n_{CO_2}\) = 0,5 mol
=>\(V_{CO_2}\) = 0,5 .22,4 = \(11,2\left(l\right)\)
\(n_{O_2}\) = \(0,2\left(mol\right)\)
=>\(V_{O_2}\) = 0,2 . 22,4 = \(4,48\left(l\right)\)
b) Ta có:
\(n_{NH_3}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\\ n_{O_2}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ \Sigma n_{hh}=1+0,5=1,5\left(mol\right)\\ V_{hh}=1,5.22,4=33,6\left(l\right)\)
