Câu hỏi của Nguyễn Trang Nhung

Question

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乇尺尺のﾚ · Accepted Answer

$a)M_X=3,75.16=60g/mol\ m_C=12\cdot\dfrac{4,48}{22,4}=2,4g\ m_H=2\cdot\dfrac{3,6}{18}=0,4g\ m_H+m_C=2,4+0,4=2,8g< m_X\ m_O=6-2,8=3,2g$

Trong X có C H O

Gọi CTPT X $C_xH_yO_z$

Ta có

$\dfrac{12x}{2,4}=\dfrac{y}{0,4}=\dfrac{16z}{3,2}=\dfrac{60}{6}\ \Rightarrow x=2;y=4;z=2$ư

Vậy CTPT X $CH_2COOH$

$b)n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\ n_{CaCO_3}=n_{CO_2}=0,2mol\ a=m_{CaCO_3}=0,2.100=20g$

Câu trả lời:

$a)M_X=3,75.16=60g/mol\ m_C=12\cdot\dfrac{4,48}{22,4}=2,4g\ m_H=2\cdot\dfrac{3,6}{18}=0,4g\ m_H+m_C=2,4+0,4=2,8g< m_X\ m_O=6-2,8=3,2g$

Trong X có C H O

Gọi CTPT X $C_xH_yO_z$

Ta có

$\dfrac{12x}{2,4}=\dfrac{y}{0,4}=\dfrac{16z}{3,2}=\dfrac{60}{6}\ \Rightarrow x=2;y=4;z=2$ư

Vậy CTPT X $CH_2COOH$

$b)n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\ n_{CaCO_3}=n_{CO_2}=0,2mol\ a=m_{CaCO_3}=0,2.100=20g$