CO + CuO → Cu + CO2 (1)
SO2 + CuO → X
\(n_{CO_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)
\(n_{hhkhí}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo Pt1: \(n_{CO}=n_{CO_2}=0,075\left(mol\right)\)
\(\Rightarrow V_{CO}=0,075\times22,4=1,68\left(l\right)\)
\(\Rightarrow\%V_{CO}=\dfrac{1,68}{6,72}\times100\%=25\%\)
\(n_{SO_2}=0,3-0,075=0,225\left(mol\right)\)
\(\Rightarrow V_{SO_2}=0,225\times22,4=5,04\left(l\right)\)
\(\Rightarrow\%V_{SO_2}=\dfrac{5,04}{6,72}\times100\%=75\%\)
PTHH: \(CO+CuO-^{t^o}\rightarrow Cu+CO2\)
\(2SO_2+2CuO+O_2\rightarrow2CuSO_4\)
=> Sau p/ứ khí thoát ra là CO2
\(n_{CO2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)
\(n_{CO2}=n_{CO}=0,075\left(mol\right)\)
\(\Rightarrow V_{CO}=V_{CO2}=1,68\left(l\right)\)
\(\Rightarrow V_{SO2}=6,72-1,68=5,04\left(l\right)\)
\(\%V_{CO}=\dfrac{1,68}{6,72}.100=25\%\)
\(\%V_{SO2}=100\%-25\%=75\%\)
- cho hỗn hợp khí gồm CO và SO2 đi qua CuO dư thì chỉ có CO phản ứng :
CuO + CO \(^{to}\rightarrow\) Cu + CO2 \(\uparrow\)
........0,075<------------0,075 mol
nCO2 = \(\dfrac{1,68}{22,4}\) = 0,075 mol
nhỗn hợp khí = \(\dfrac{6,72}{22,4}\) = 0,3 mol
=>%VCO = \(\dfrac{0,075}{0,3}\) .100% = 25%
=>%VSO2 = 100 % - 25 % = 75 %