Còn câu này nữa thui, giúp mik mn ;-;

$n_{NaOH}= 0,025(mol)\\n_{O_2}=0,2675(mol)\\n_{CO_2}=0,24(mol)\\n_{COO}= n_{NaOH}=0,025(mol)$ 

0,075 mol T chứa a mol COO

Đốt 0,075 mol T thu được b mol $H_2O$ 

$BTNT.O:\\2a + 0,2675. 2 = b + 0,24.2(1)\\BTKL:\\m_T = 18b + 0,24 . 44-0,2675 . 32=18b +2\\\dfrac{n_{COO}}{m_T}=\dfrac{a}{18b+2}= \dfrac{0,025}{2,395}(2)\\(1)(2)→ a = 0,05(mol), b= 0,155(mol)\\T: C_xH_yO_z\\x = \dfrac{0,24}{0,075}=3,2\\y = \dfrac{0,155. 2}{0,075}=\dfrac{62}{15}\\z = \dfrac{0,05. 2}{0,075}={4}{3}\\→COO= \dfrac{2}{3}\\k = \dfrac{2x - y + 2}{2}=\dfrac{32}{15}\\n_{Br_2} = n_{pi} = 0,075(\dfrac{32}{15} - \dfrac{2}{3}) = 0,11(mol)\\m_T = 18b +2=4,79(g)$

Tam xuất:

$p = \dfrac{3,832. 0,11}{4,79}= 0,088(mol)$