Còn thiếu nguyên tử lưu huỳnh nha bn,

Bn ghi rõ đề vào nha.

\(M_H=98-32-16.4=2\left(đvC\right)\)

 ⇒ Có 2:1 = 2 ntử H trong 1 ptử H₂SO₄

Gọi CTHH của axit sunfuric là: H_xSO₄

Ta có: \(M_{H_xSO_4}=98\left(đvC\right)\)

Mà \(M_{H_xSO_4}=M_H.x+M_S+M_O.4=98\left(đvC\right)\)

\(M_{H_xSO_4}=1.x+32+16.4=98\left(đvC\right)\)

\(M_{H_xSO_4}=x+32+64=98\left(đvC\right)\)

=> x = 2 

=> CTHH của axit sunfuric là: H₂SO₄

cái này là hoá bạn đừng đăng vào online math bạn đăng câu hỏi vào link này nhé:

https://h.vn/

@Út Nhỏ Jenny: làm gì có link này c nhỉ???

\(a)Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b)Đặt:n_{Zn}=x\left(mol\right);n_{Fe}=y\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}65x+56y=12,1\\x+y=0,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\\ \Rightarrow m_{Fe}=0,1.56=5,6\left(g\right);m_{Zn}=0,1.65=6,5\left(g\right)\\ c)n_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow C\%_{H_2SO_{\text{ 4}}}=\dfrac{0,2.98}{196}.100=10\%\)

Coi $n_{H_2O} = 1(mol)$

Suy ra : $n_H = 2(mol)$

$\Rightarrow n_O = 2.0,875 = 1,75(mol)$

Ta có : $n_O = 4n_{H_2SO_4} + n_{H_2O}$

$\Rightarrow n_{H_2SO_4} = 0,1875(mol)$

$m_{dd} = m_{H_2SO_4} + m_{H_2O} = 0,1875.98 + 1.18 = 36,375(gam)$

$C\%_{H_2SO_4} = \dfrac{0,1875.98}{36,375}.100\% = 50,5\%$

\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.20\%}{98}=0,204\left(mol\right)\)

PTHH:

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,04      0,04           0,04    

\(\dfrac{0,04}{1}< \dfrac{0,204}{1}\) --> H2SO4 dư

\(C\%_{CuSO_4}=\dfrac{0,04.160}{3,2+100}.100\%=6,2\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{3,2+100}.100\%=19\%\)

\(n_{MgO}=\dfrac{0,8}{40}=0,02\left(mol\right)\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(n_{H_2SO_4}=n_{MgO}=0,02\left(mol\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,02.98}{50}.100\%=3,92\%\)

\(n_{MgO}=\dfrac{0,8}{40}=0,02\left(mol\right)\)

PTHH :

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

0,02       0,02

\(a,m_{H_2SO_4}=0,02.98=1,96\left(g\right)\)

\(C\%=\dfrac{1,96}{50}.100\%=3,92\%\)

\(n_{MgO}=\dfrac{0,8}{40}=0,02\left(mol\right)\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(n_{MgO}=n_{H2SO4}=0,02\left(mol\right)\)

\(C\%_{ddH2SO4}=\dfrac{0,02.98}{50}.100\%=3,92\%\)

Câu 1:

CuO + H₂SO₄ → CuSO₄ + H₂O

\(n_{CuO}=\frac{3,2}{80}=0,04\left(mol\right)\)

\(m_{H_2SO_4}=200\times9,8\%=19,6\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\frac{19,6}{98}=0,2\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{H_2SO_4}\)

Theo bài: \(n_{CuO}=\frac{1}{5}n_{H_2SO_4}\)

Vì \(\frac{1}{5}< 1\) ⇒ H₂SO₄ dư

Dung dịch sau pư gồm: H₂SO₄ dư và CuSO₄

Ta có: \(m_{dd}saupư=3,2+200=203,2\left(g\right)\)

Theo Pt: \(n_{H_2SO_4}pư=n_{CuO}=0,04\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}dư=0,2-0,04=0,16\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}dư=0,16\times98=15,68\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}dư=\frac{15,68}{203,2}\times100\%=7,72\%\)

Theo Pt: \(n_{CuSO_4}=n_{CuO}=0,04\left(mol\right)\)

\(\Rightarrow m_{CuSO_4}=0,04\times160=6,4\left(g\right)\)

\(\Rightarrow C\%_{CuSO_4}=\frac{6,4}{203,2}\times100\%=3,15\%\)

Câu 2:

ZnO + H₂SO₄ → ZnSO₄ + H₂O

\(n_{ZnO}=\frac{8,1}{81}=0,1\left(mol\right)\)

\(m_{H_2SO_4}=200\times24,5\%=49\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\frac{49}{98}=0,5\left(mol\right)\)

Theo Pt: \(n_{ZnO}=n_{H_2SO_4}\)

Theo bài: \(n_{ZnO}=\frac{1}{5}n_{H_2SO_4}\)

Vì \(\frac{1}{5}< 1\) ⇒ H₂SO₄ dư

Dung dịch sau pư gồm: H₂SO₄ dư và ZnSO₄

Ta có: \(m_{dd}saupư=8,1+200=208,1\left(g\right)\)

Theo PT: \(n_{H_2SO_4}pư=n_{ZnO}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}dư=0,5-0,1=0,4\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,4\times98=39,2\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\frac{39,2}{208,1}\times100\%=18,84\%\)

Theo pT: \(n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\)

\(\Rightarrow m_{ZnSO_4}=0,1\times161=16,1\left(g\right)\)

\(\Rightarrow C\%_{ZnSO_4}=\frac{16,1}{208,1}\times100\%=7,74\%\)