Mn giúp mk bài hoá vs ạ Camon

Đáp án:

$\rm n_{C_6H_5OH}=\dfrac{9,5}{94}≈0,1\ (mol)$

Phản ứng xảy ra:

$\rm C_6H_5OH+NaOH\to C_6H_5ONa+H_2O$

Ta có: $\rm n_{C_6H_5ONa}=n_{C_6H_5OH}=0,1\ (mol)$

$\rm\to m=0,1×116=11,6\ (g)$

`n_{C_6H_5OH}={9,5}/{94}={19}/{188}(mol)`

`C_6H_5OH+NaOH->C_6H_5ONa+H_2O`

Theo PT: `n_{C_6H_5ONa}=n_{C_6H_5OH}={19}/{188}(mol)`

`->m={19}/{118}.116\approx 11,72(g)`