a)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

b)

- Cu(OH)₂:

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)

- Fe(OH)₃:

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\)

- Fe(OH)₂:

\(2H_2O\underrightarrow{đp}2H_2+O_2\)

\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

Tổng số hạt trong nguyên tử `X` là `50=>2e+n=50` `(p=e)`   `(1)`

Vì số hạt ko mang điện tích nhiều hơn số hạt mang điện tích âm là `2`

             `=>n-e=2`  `(2)`

Từ `(1);(2)=>{(e=16=p),(n=18):}`

Vậy nguyên tử `X` là Lưu huỳnh `(S)` có `p=e=16` hạt; `n=18` hạt

Vì tổng số hạt của nguyên tử `X` là `126=>2e+n=126` `(p=e)`   `(1)`

Vì số nơtron nhiều hơn electron `12` hạt `=>n-e=12`  `(2)`

Từ `(1);(2)=>{(e=38=p),(n=50):}`

   `->` Nguyên tử `X` là `Sr` `(Stronti)`

\(Tacó:\left\{{}\begin{matrix}P=E\\P+E+N=126\\N-E=12\end{matrix}\right.\\ \Rightarrow P=38,N=50\\ P=38\Rightarrow XlàStronti\left(Sr\right)\)

\(PTHH:MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4dư}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.\dfrac{100.24\%}{40}=0,3\left(mol\right)\\ \Rightarrow n_{H_2SO_4pứ}=1-0,3=0,7\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{MgO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\\ Tacó:40x+160y=32\\ Mặckhác:x+3y=0,7\\ \Rightarrow x=0,4;y=0,1\\ \Rightarrow\%MgO=\dfrac{0,4.40}{32}.100=50\%;\%Fe_2O_3=50\%\)

\(CaCO3\rightarrow CaO+CO2 \)

2mol       

\(nCaO=90\%.2=1,8\left(mol\right)\Rightarrow mCaO=1,8.56=100,8\left(g\right)\)

\(nCO2=1,8\left(mol\right)\Rightarrow mCO2=1,8.44=79,2\left(g\right)\)

tính thể tích CO2 thu dc sau phản ứng 

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ n_{Fe_2\left(SO_4\right)_3}=\dfrac{12}{400}=0,03\left(mol\right)\\ n_{Fe_2O_3}=n_{Fe_2\left(SO_4\right)_3}=0,03\left(mol\right)\\ \Rightarrow m_{Fe_2O_3}=0,03.160=4,8\left(g\right)\\ n_{H_2SO_4}=3n_{Fe_2O_3}=0,09\left(mol\right)\\ C\%_{H_2SO_4}=\dfrac{0,09.98}{300}.100=2,94\%\)

a) \(n_{NaOH}=\dfrac{150.20\%}{40}=0,75\left(mol\right)\); \(n_{CuSO_4}=\dfrac{200.20\%}{160}=0,25\left(mol\right)\)

PTHH: \(2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)
Xét tỉ lệ: \(\dfrac{0,75}{2}>\dfrac{0,25}{1}\) => NaOH dư, CuSO₄ hết

PTHH: \(2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

               0,5<-------0,25-------->0,25--------->0,25

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,25.98=24,5\left(g\right)\)

b)

m_{dd sau pư} = 150 + 200 - 24,5 = 325,5 (g)

\(\left\{{}\begin{matrix}C\%_{NaOH\left(dư\right)}=\dfrac{40\left(0,75-0,5\right)}{325,5}.100\%=3,07\%\\C\%_{Na_2SO_4}=\dfrac{0,25.142}{325,5}.100\%=10,91\%\end{matrix}\right.\)

Tham khảo :

\(a,n_{NaOH}=\dfrac{150.20\%}{40}=0,75\left(mol\right)\)

\(n_{CuSO_4}=\dfrac{200.20\%}{160}=0,25\left(mol\right)\)

\(2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

\(0,75---\rightarrow0,25\)

Lập tỉ lệ :

\(\dfrac{0,75}{2}>\dfrac{0,25}{1}\Rightarrow\) sau phản ứng NaOH dư

\(n_{Cu\left(OH\right)_2}=n_{CuSO_4}\)

\(m_{Cu\left(OH\right)_2}=0,25.98=24,5\left(g\right)\)

Dung dịch sau phản ứng gồm Na2SO4 và NaOH dư

\(m_{NaOH\left(dư\right)}=\left(0,75-0,25.2\right).40=10\left(g\right)\)

\(m_{Na_2SO_4}=0,25.142=35,5\left(g\right)\)

a)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

b)

- Cu(OH)₂:

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)

- Fe(OH)₃:

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\)

- Fe(OH)₂:

\(2H_2O\underrightarrow{đp}2H_2+O_2\)

\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+2NaCl\)

a. \(Na_2O+H_2O\rightarrow2NaOH\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

b. \(CuO+H_2O\rightarrow Cu\left(OH\right)_2\)

\(Fe_2O_3+H_2O\rightarrow2Fe\left(OH\right)_2\)

\(a.Đặt:\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{Al_2O_3}=y\left(mol\right)\\n_{FeO}=z\left(mol\right)\end{matrix}\right.\\ Tacó:80x+102y+72z=47,8\left(g\right)\left(1\right)\\ \\ Phần1:\left\{{}\begin{matrix}n_{CuO}=\dfrac{x}{2}\left(mol\right)\\n_{Al_2O_3}=\dfrac{y}{2}\left(mol\right)\\n_{FeO}=\dfrac{z}{2}\left(mol\right)\end{matrix}\right.\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ FeO+2HCl\rightarrow FeCl_2+H_2O\\ n_{HCl}=\dfrac{x}{2}.2+\dfrac{y}{2}.6+\dfrac{z}{2}.2=\dfrac{400.8,2125\%}{36,5}=0,9\left(mol\right)\left(2\right)\\ Phần2:\\ CuO+H_2SO_{4\left(đ,n\right)}\rightarrow CuSO_4+H_2O\\ Al_2O_3+3H_2SO_{4\left(đ,n\right)}\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ 2FeO+4H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\\ n_{FeO}=\dfrac{z}{2}=2n_{SO_2}=2.\dfrac{1,68}{22,4}=0,15\\ \Rightarrow z=0,3\left(mol\right)\left(3\right)\\ Từ\left(1\right),\left(2\right),\left(3\right)\Rightarrow\left\{{}\begin{matrix}x=0,126\\y=0,158\\z=0,3\end{matrix}\right.\\ \Rightarrow\%CuO=21,09,\%Al_2O_3=33,72\%,\%FeO=45,19\%\\ b.m_{ddB}=\dfrac{47,8}{2}+400=123,9\left(g\right)\\ n_{CuCl_2}=\dfrac{0,126}{2}=0,063\left(mol\right)\\ n_{AlCl_3}=\dfrac{0,158}{2}.2=0,158\left(mol\right)\\ n_{FeCl_2}=\dfrac{0,3}{2}=0,15\left(mol\right)\\ C\%_{CuCl_2}=\dfrac{0,063.135}{123,9}.100=6,86\%\\ C\%_{AlCl_3}=\dfrac{0,158.133.5}{123,9}.100=17,02\%\\ C\%_{FeCl_2}=\dfrac{0,15.127}{123,9}.100=15,38\%\)