Đáp án:

`↓`

Giải thích các bước giải:

Bài 6:

`a)` $n_{Cl_2(đktc)}$`=``\frac{V}{22,4}``=``\frac{3,36}{22,4}``=``0,15` `(mol)`

`PTHH`             `2NaOH``+``Cl_2``→``NaClO``+``NaCl``+``H_2O`

                           `0,3`        `0,15`        `0,15`       `0,15`                     `mol`

`→``n_{NaOH}``=``n_{Cl_2}``=``\frac{0,15.2}{1}``=``0,3` `(mol)`

`→``V_{NaOH}``=``\frac{n}{C_M}``=``\frac{0,3}{2}``=``0,15` `(l)`

`b)` `V_{sau}``=``V_{NaOH}``=``0,15` `(l)`

Theo pt, ta có: `n_{NaClO}``=``n_{Cl_2}``=``0,15` `(mol)`

                        `n_{NaCl}``=``n_{Cl_2}``=``0,15` `(mol)`

`→`$C_{M_{NaClO}}$`=``\frac{n}{V_{sau}}``=``\frac{0,15}{0,15}``=``1` `(M)`

`→`$C_{M_{NaCl}}$`=``\frac{n}{V_{sau}}``=``\frac{0,15}{0,15}``=``1` `(M)`

Bạn tham khảo!

Đáp án:

$a) 0,15l/150ml$

$b)$

$+)$ $CM_{NaCl}$ $=1M$

$+)$ $CM_{NaClO}$ $=1M$

Giải thích các bước giải:

Ta có PTHH sau:

$2NaOH+Cl_2$ $\rightarrow$ $NaCl+NaClO+H_2O$ 

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Có $n_{Cl_2}$ $=$ $\dfrac{3,36}{22,4}$ $=0,15$ mol

Dựa vào PTHH) $n_{NaOH}$ $=0,15.2=0,3$ mol

a) Vậy $V_{NaOH}$ $=$ $\dfrac{n}{CM}$ $=$ $\dfrac{0,3}{2}$ $=0,15l=150ml$

b) Dựa vào PTHH) $n_{NaCl}$ $=$ $n_{NaClO}$ $=$ $n_{Cl_2}$ $=0,15$ mol

Mà $V_{\text{sau}}$ $=$ $Vdd_{\text{đầu}}$ $=$ $V_{NaOH}$ $=0,15l$

Vậy $CM_{NaCl}$ $=$ $\dfrac{0,15}{0,15}$ $=1M$

Vậy $CM_{NaClO}$ $=$ $\dfrac{0,15}{0,15}$ $=1M$