Question

1) Dot chay hoan toan 7,54g hon hop CH₄, C₂H₆, C₃H₆, C₃H₈ va C₄H₁₀ duoc hon hop X. Dan X qua dung dich NaOH dam dac, du nhan thay binh tang them 34,58g. Neu dot chay hoan toan 50,00g hon hop tren thi thai ra moi turong bao nhieu gam khi CO₂

Answer 1

Coi hh gồm C và H.

Gọi: \(\left\{{}\begin{matrix}n_C=x\left(mol\right)\\n_H=y\left(mol\right)\end{matrix}\right.\) ⇒ 12x + y = 7,54 (g)

BTNT C, có: n_CO2 = n_C = x (mol)

BTNT H, có: \(n_{H_2O}=\dfrac{1}{2}n_H=\dfrac{1}{2}y\left(mol\right)\)

Mà: m _{bình NaOH tăng} = m_CO2 + m_H2O

⇒ 34,58 = 44x + 1/2y.18 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,52\left(mol\right)\\y=1,3\left(mol\right)\end{matrix}\right.\)

→ Trong 50 (g) hh có: \(n_C=0,52.\dfrac{50}{7,54}=\dfrac{100}{29}\left(mol\right)=n_{CO_2}\)

\(\Rightarrow m_{CO_2}=\dfrac{100}{29}.44=\dfrac{4400}{29}\left(g\right)\)