Question

 cho 2, 464 lít CO2 đktc đi vào dung dịch NaOH sinh ra 11,44g hỗn hợp 2 muối là NaHCO3 và Na2CO3 .xác đinh khối lượng của mỗi muối.

Answer 1

nCO2=0,11(mol" role="presentation" style="background:transparent; border:0px; color:rgb(0, 0, 0); direction:ltr; display:inline; float:none; font-family:arial,liberation sans,dejavu sans,sans-serif; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-wrap:normal" class="MathJax">nCO2=0,11(molCO2+NaOH→NaHCO3" role="presentation" style="background:transparent; border:0px; direction:ltr; display:inline; float:none; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">CO2+NaOH→NaHCO3x" role="presentation" style="background:transparent; border:0px; direction:ltr; display:inline; float:none; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">x                                     x       (mol)CO2+2NaOH→Na2CO3+H2O" role="presentation" style="background:transparent; border:0px; direction:ltr; display:inline; float:none; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">CO2+2NaOH→Na2CO3+H2Oy" role="presentation" style="background:transparent; border:0px; direction:ltr; display:inline; float:none; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">y                                         y      (mol){{⇒{mNaHCO3=0,01.84=0,84(g)mNa2CO3=0,1.106=10,6(g)" role="presentation" style="background:transparent; border:0px; direction:ltr; display:inline; float:none; font-size:13.696px; line-height:normal; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:0px; position:relative; vertical-align:baseline; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax">⇒{mNaHCO3=0,01.84=0,84(g)mNa2CO

Answer 2

nCO2 = 0,11 (mol)

CO2 + NaOH => NaHCO3

x                          x (mol)

CO2 + 2NaOH => NA2CO3 + H2O

y                               y (mol)

Theo đề ta có :

x + y = 0,11      

84x + 106y = 11,44 

<=> x = 0,11

<=> y = 0,1

mNaOHCO3 = 0,01 . 84 = 0,84 (g)

mNaO2CO3 = 0,1 . 106 = 10,6 (g)

Answer 3

\(nCO2=0,11\left(mol\right)\)

\(CO2+NaOH\rightarrow NaHCO3\)

 \(x\)            \(x\)     \(\left(mol\right)\)

\(CO2+2NaOH\rightarrow Na2CO3+H2O\)

\(y\)                                \(y\)     \(\left(mol\right)\)

Theo bài ra ta có : \(\begin{cases}x+y=0,11\\84x+106y=11,44\end{cases}\Leftrightarrow}\begin{cases}x=0,01\\y=0,1\end{cases}\)

\(\Rightarrow\begin{cases}mNaHCO3=0,01.84=0,84\left(g\right)\\mNa2CO3=0,1.106=10,6\left(g\right)\end{cases}\)