a)
\(n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ n_{CO_2}=\dfrac{m}{M}=\dfrac{4,4}{12+16\cdot2}=0,1\left(mol\right)\\ n_{Ca\left(OH\right)_2}=\dfrac{m}{M}=\dfrac{7,4}{40+\left(16+1\right)\cdot2}=0,1\left(mol\right)\\ n_{N_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
b)
\(m_{Cu}=n\cdot M=0,1\cdot64=6,4\left(g\right)\\ m_{NaCl}=n\cdot M=0,3\cdot\left(23+35,5\right)=17,55\left(g\right)\\ m_{Na_2CO_3}=n\cdot M=0,06\cdot\left(23\cdot2+12+16\cdot3\right)=6,36\left(g\right)\)
c)
\(V_{CO_2\left(dktc\right)}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\\ V_{SO_2\left(dktc\right)}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\\ V_{CH_4}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\\ n_{N_2O}=\dfrac{m}{M}=\dfrac{4,4}{14\cdot2+16}=0,1\left(mol\right)=>V_{N_2O\left(dktc\right)}=n\cdot22,4=0,1\cdot22,4=2,24\left(l\right)\)
