a) nFe= \(\frac{m_{Fe}}{M_{Fe}}=\frac{5,6}{56}=0,1\left(mol\right)\)
nCu=\(\frac{m_{Cu}}{M_{Cu}}=\frac{64}{64}=1\left(mol\right)\)
nAl= \(\frac{m_{Al}}{M_{Al}}=\frac{27}{27}=1\left(mol\right)\)
b) \(n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{44}{44}=1\left(mol\right)\)
\(n_{H_2}=\frac{m_{H_2}}{M_{H_2}}=\frac{4}{2}=2\left(mol\right)\)
a) nFe = 5,6/56 = 0,1 mol
nCu = 64/64 = 1 mol
nAl = 27/27 = 1 mol
b) nCO2 = 44/44 = 1 mol
=> VCO2 = 1.22,4 = 22,4 l
nH2 = 4/2 = 2 mol
=> VH2 = 2.22,4 = 44,8 l
a) nFe = \(\frac{5.6}{56}=0.1\)(mol)
nCu = \(\frac{64}{64}=1\) (mol)
nAl = \(\frac{27}{27}=1\) (mol)
b) nCO2 = \(\frac{44}{44}=1\)(mol)
nH2 = \(\frac{4}{2}=2\) (mol)
\(\Rightarrow\)nhh = 1 + 2 = 3 (mol)
\(\Rightarrow\)Vhh = 3 * 22.4 = 67.2 (l)