\(n_{HCl}=0,2.3,5=0,7\left(mol\right)\\ n_{CuO}=a;n_{Fe_2O_3}=b\\ CuO+2HCl\xrightarrow[]{}\Rightarrow CuCl_2+H_2O\\ Fe_2O_3+6HCl\xrightarrow[]{}2FeCl_3+H_2O\\ \left\{{}\begin{matrix}2a+6b=0,7\\80a+160b=20\end{matrix}\right.\\\Rightarrow a=0,05;b=0,1\\ \%_{CuO}=\dfrac{0,05.80}{20}\cdot100=20\%\\ \%_{Fe_2O_3}=\dfrac{0,1.160}{20}\cdot100=80\%\)

\(\left(1\right)CaO+CO_2\xrightarrow[]{}CaCO_3\\ \left(2\right)CaO+H_2O\xrightarrow[]{}Ca\left(OH\right)_2\\ \left(3\right)Ca\left(OH\right)_2+CO_2\xrightarrow[]{}CaCO_3+H_2O\\ \left(4\right)CaCO_3\xrightarrow[]{t^0}CaO+CO_2\\ \left(5\right)CaO+2HCl\xrightarrow[]{}CaCl_2+H_2O\)

\(a)CO_2+2NaOH\xrightarrow[]{}Na_2CO_3+H_2O\\ b)n_{CO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ n_{NaOH}=0,05.2=0,1\left(mol\right)\\ C_{M_{NaOH}}=\dfrac{0,1}{0,1}=1\left(M\right)\)

Gọi số mol của 440 g chất béo là b (mol)

\(\left(RCOO\right)_3C_3H_5+3NaOH->3RCOONa+C_3H_5\left(OH\right)_3\\ b=\dfrac{60}{40\cdot3}=0,5mol\\ M_{cb}=\dfrac{440}{0,5}=880\left(g\cdot mol^{-1}\right)\\ C_{cb}=\dfrac{28,5}{0,5}=57\\ CB:C_{57}H_{100}O_6,hay:C_nH_{2n-4-2k}O_6\left(k:số.lk.\pi.C=C.hoặc.C\equiv C\right)\\ 2.57-4-2k=100\\ k=5\\ n_{H_2}=kb=2,5mol\\ m_{H_2}=2,5.2=5g\)

 Đồng đẳng hóa:

\(\left\{{}\begin{matrix}\left(HCOO\right)_3C_3H_5:a\\CH_2:b\\H_2:-c\end{matrix}\right.\) 

Ta có: \(\left\{{}\begin{matrix}m=176a+14b-2c=440\\n_{NaOH}=3a=\dfrac{60}{40}=1,5\\\dfrac{n_{cbeo}}{n_{CO_2}}=\dfrac{0,5}{28,5}=\dfrac{a}{6a+b}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=25,5\\c=2,5\end{matrix}\right.\)

Vậy \(m_{H_2}=2,5.2=5\left(g\right)\)

Từ đề ra:

\(2p+n=40\\ 2p-n=12\\ p=\dfrac{40+12}{4}=13\\ 2p=26\)

Vậy X có 26 hạt mang điện

p = e

Tổng hạt = p + n + e = 2p + n

Hạt mang điện = p + e = 2p