A

\(V_{CO_2}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\\ n_{O_2}=n\cdot22,4=0,25\cdot22,4=5,6\left(l\right)\)

B

\(n_{N_2}=\dfrac{m}{M}=\dfrac{21}{14\cdot2}=0,75\left(mol\right)\\ V_{N_2}=n\cdot22,4=0,75\cdot22,4=16,8\left(l\right)\)

\(n_{CO_2}=\dfrac{m}{M}=\dfrac{8,8}{12+16\cdot2}=0,2\left(mol\right)\\ V_{CO_2}=n\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\)

C

\(n_{H_2}=\dfrac{9\cdot10^{23}}{6\cdot10^{23}}=1,5\left(mol\right)\\ V_{H_2}=n\cdot22,4=1,5\cdot22,4=33,6\left(l\right)\)

\(n_{CO_2}=\dfrac{0,3\cdot10^{23}}{6\cdot10^{23}}=0,05\left(mol\right)\\ V_{CO_2}=n\cdot22,4=0,05\cdot22,4=1,12\left(l\right)\)

a) \(n_C=\dfrac{4}{12}=\dfrac{1}{2}\left(mol\right)\)

\(n_P=\dfrac{62}{31}=2\left(mol\right)\)

\(n_{Fe}=\dfrac{42}{56}=0,75\left(mol\right)\)

b) \(n_{H2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\)

\(n_{CO2}=\dfrac{95,48}{44}=2,17\left(mol\right)\)

\(n_{NaCl}=\dfrac{14,625}{58,5}=0,25\left(mol\right)\)

a.mCuSO4=0.6×160=96g

b.nCl2=3×10^23/6×10^23=0.5mol

mCl2=0.5×71=35.5g

c.nCH4=0.5mol

mCH4=0.5×16=8g

d.mC6H12O6=180×1.5=270g

PTHH : SO₂ + Ba(OH)₂ -> BaSO₃ \(\downarrow\)+ H₂O 

            \(\dfrac{197}{2330}\)                         \(\dfrac{197}{2330}\)

\(n_{Ba\left(OH\right)_2}=2.0,1=0,2\left(mol\right)\)

\(n_{BaSO_4}=\dfrac{19.7}{233}=\dfrac{197}{2330}\left(mol\right)\)

Ta thấy : Ba(OH)₂ dư

\(m_{SO_2}=\dfrac{197}{2330}.64=\dfrac{6304}{1165}\left(g\right)\)

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