BT3
\(n_{CO_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\)
\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
tl1..............2....................1..........1.(Mol)
Br0,07..........................0.07.............(mol)
\(m_{Na_2CO_3}=n.M=0,07.106=7,42\left(mol\right)\)
BT4
\(n_{CO_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\)
\(n_{NaOH}=\dfrac{3,2}{40}=0,08\left(mol\right)\)
\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
tl1...............2...................1..........1.(mol)
br 0,04..... 0,08..........0,04............(mol)
SO sánh:\(\dfrac{n_{CO_2}}{1}>\dfrac{n_{NaOH}}{2}\left(\dfrac{0,07}{1}>\dfrac{0,08}{2}\right)\)
=>CO2 dư tính theo mol NaOH
=>\(m_{Na_2CO_3}=n.M=0,04.106=4,24\left(g\right)\)
