Tham khảo :
$$n_{phenol}=\dfrac{11,28}{94}=0,12(mol)$$
$$n_{HNO_3}=\dfrac{27.63}{63.100}=0,27(mol)$$
- Vì $$\dfrac{n_{HNO_3}}{3}
-> Tính theo $$HNO_3$$
$$C_6H_5OH+3HNO_3 \xrightarrow{t^o,H_2SO_4(đ)}C_6H_2(NO_2)_3OH+3H_2O$$
- Hiệu suất $$100\%:$$
$$m_{axit \,\, picric}=\dfrac{0,27}{3}.229=20,61(gam)$$
 

C_6H_5OH+ 3HNO_3$$\xrightarrow{t°,H_2SO_4}$$ C_6H_2(NO_2)_3OH + 3H_2O 
n_{C_6H_5OH}={11,28}/{94}=0,12(mol)
n_{HNO_3}= {63%. 27}/{63}= 0,27(mol) 
n_{C_6H_5OH}> 1/3n_{HNO_3} 
-> HNO_3 hết, phenol dư 
Theo pt: 
n_{C_6H_2(NO_2)_3OH}= 1/3n_{HNO_3}= 0,09(mol)
m_{C_6H_2(NO_2)_3OH}= 0,09. 229= 20,61(g)