Đáp án:

$a)C_{M_{dd bazo}}=1M$

$b)V_{dd_{H_2SO_4}}\approx 53,73(ml)$

Giải thích các bước giải:

$a)PTHH:Na_2O+H_2O\to 2NaOH$

$n_{Na_2O}=\dfrac{7,75}{62}=0,125(mol)$

Theo PT: $n_{NaOH}=2n_{Na_2O}=2.0,125=0,25(mol)$

Đổi $250ml=0,25l$

$\Rightarrow C_{M_{NaOH}}=\dfrac{0,25}{0,25}=1M$

$b)PTHH:2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$

Theo PT: $n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.0,25=0,125(mol)$

$\Rightarrow m_{H_2SO_4}=0,125.98=12,25(g)$

$\Rightarrow m_{dd_{H_2SO_4}}=\dfrac{12,25}{20\%}=61,25(g)$

$\Rightarrow V_{dd_{H_2SO_4}}=\dfrac{m_{dd}}{D}=\dfrac{61,25}{1,14}\approx 53,73(ml)$

Đáp án:

`↓`

Giải thích các bước giải:

Bài 7:

`a)` Đổi `250``ml``=``0,25``l`

`n_{Na_2O}``=``\frac{m}{M}``=``\frac{7,75}{62}``=``0,125` `(mol)`

`PTHH`              `Na_2O``+``H_2O``→``2NaOH`

                         `0,125`                         `0,25`          `mol`

`→``n_{NaOH}``=``n_{Na_2O}``=``\frac{0,125.2}{1}``=``0,25` `(mol)`

`→`$C_{M_{NaOH}}$`=``\frac{n}{V}``=``\frac{0,25}{0,25}``=``1` `(M)`

`b)` 

`PTHH`               `H_2SO_4``+``2NaOH``→``Na_2SO_4``+``2H_2O`

                             `0,125`          `0,25`                                                 `mol`

`→``n_{H_2SO_4}``=``n_{NaOH}``=``\frac{0,25.1}{2}``=``0,125` `(mol)`

`→``m_{H_2SO_4}``=``n``.``M``=``0,125``.``98``=``12,25` `(g)`

`⇒`$m_{dd_{H_2SO_4}}$`=``\frac{m.100%}{C%}``=``\frac{12,25.100%}{20%}``=``61,25` `(g)`

`⇒``V_{H_2SO_4}``=``\frac{m_{dd}}{D}``=``\frac{61,25}{1,14}``≈``53,73` `(ml)`