\(a.\)
\(n_{CO_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(CuO+CO\underrightarrow{^{^{t^0}}}Cu+CO_2\)
Cách 1 :
\(n_{CO}=n_{CO_2}=0.2\left(mol\right)\)
\(\Rightarrow V_{CO}=0.2\cdot22.4=4.48\left(l\right)\)
Cách 2 :
\(n_{CuO}=n_{Cu}=n_{CO_2}=0.2\left(mol\right)\)
Bảo toàn khối lượng :
\(m_{CuO}+m_{CO}=m_{Cu}+m_{CO_2}\)
\(\Rightarrow m_{CO}=64\cdot0.2+0.2\cdot44-0.2\cdot80=5.6\left(g\right)\)
\(n_{CO}=\dfrac{5.6}{28}=0.2\left(mol\right)\)
\(V_{CO}=4.48\left(l\right)\)
\(b.\)
\(n_{CO}=n_{CO_2}=\dfrac{1.1}{44}=0.025\left(mol\right)\)
\(\Rightarrow V_{CO}=0.025\cdot22.4=0.56\left(l\right)\)
a)
$CuO + CO \xrightarrow{t^o} Cu + CO_2$
Theo PTHH :
$n_{CO} = n_{CO_2} \Rightarrow V_{CO\ pư} = V_{CO_2} = 4,48(lít)$
b)
$n_{CO} = n_{CO_2} = \dfrac{1,1}{44} = 0,025(mol)$
$V_{CO} = 0,025.22,4 = 0,56(lít)$
