\(n_{FeO}=\dfrac{10,8}{72}=0,15\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.20\%}{100\%}:98=\dfrac{10}{49}\left(mol\right)\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

0,15----> 0,15 -----> 0,15----->0,15

Xét \(\dfrac{10}{49}:1>\dfrac{0,15}{1}\) => axit dư.

Sau khi phản ứng kết thúc, trong dung dịch có:

\(m_{H_2SO_4}=\left(\dfrac{10}{49}-0,15\right).98=5,3\left(g\right)\)

\(m_{FeSO_4}=0,15.152=22,8\left(g\right)\)

\(m_{H_2O}=0,15.18=2,7\left(g\right)\)

đọc kỹ đề nhé ctv: từng chất trong dung dịch

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

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

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

C%cucl₂ = 2.7.100/100 ạ?

Zn + H₂SO₄ -> ZnSO₄ + H₂ (1)

n_Zn=0,1(mol)

n_H2SO4=0,2(mol)

Sau PƯ 1 ta thấy còn 0,4 mol H₂SO₄ dư

Từ 1:

n_H2=n_ZnSO4=n_Zn=0,1(mol)

C% dd ZnSO₄=\(\dfrac{161.0,1}{6,5+196-0,1.2}.100\%=8\%\)

C% dd H₂SO₄=\(\dfrac{98.0,4}{196+6,5-0,2}.100\%=19,377\%\)

sao số mol H2SO4 lại = 0, 2 vậy ạ ? anh tính chi tiết hộ em với !

Bài 1:

n_{H2SO4 bđ} = 0,5 . 0,2 = 0,1 mol

n_H2 = \(\dfrac{1,792}{22,4}=0,08\left(mol\right)\)

Pt: Zn + H₂SO₄ --> ZnSO₄ + H₂

............0,08 mol<-0,08 mol<-0,08 mol

Theo pt: n_{H2SO4 pứ} = n_H2 = 0,08 mol < 0,1 mol

=> HCl dư

C_{M H2SO4 dư} = \(\dfrac{\left(0,1-0,08\right)}{0,2}=0,1M\)

C_{M ZnSO4} = \(\dfrac{0,08}{0,2}=0,4M\)

Bài 2:

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

m_H2SO4 = \(\dfrac{150\times32,666}{100}=49\left(g\right)\)

n_H2SO4 = \(\dfrac{49}{98}=0,5\left(mol\right)\)

Pt: CuO + H₂SO₄ --> CuSO₄ + H₂O

0,04 mol->0,04 mol->0,04 mol

Xét tỉ lệ mol giữa CuO và H2SO4:

\(\dfrac{0,04}{1}< \dfrac{0,5}{1}\)

Vậy H2SO4 dư

m_{H2SO4 dư} = (0,5 - 0,04) . 98 = 45,08 (g)

m_CuSO4 = 0,04 . 160 = 6,4 (g)

mdd sau pứ = mCuO + mdd H2SO4 = 3,2 + 150 = 153,2 (g)

C% dd CuSO4 = \(\dfrac{6,4}{153,2}.100\%=4,177\%\)

C% dd H2SO4 dư = \(\dfrac{45,08}{153,2}.100\%=29,425\%\)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot29,4\%}{98}=0,6\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,6}{3}\) \(\Rightarrow\) Axit còn dư, Nhôm p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,6-0,3=0,3\left(mol\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,3\cdot98=29,4\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72 \left(l\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Al}+m_{ddH_2SO_4}-m_{H_2}=204,8\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{204,8}\cdot100\%\approx16,7\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{29,4}{204,8}\cdot100\%\approx14,36\%\end{matrix}\right.\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{H_2SO_4}=\dfrac{10.196}{100}=19,6\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)

PTHH:_______\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Theo PT:mol:___1..........1..................1.............1

Theo ĐB:mol:___0,1.......0,2...............................

\(\Rightarrow H_2SO_4\)dư,Zn pứ hết

Theo PT: \(n_{H_2SO_4pư}=n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4dư}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,1.98=9,8;m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{9,8}{196+6,5}.100\%\approx4,84\%\)

\(\Rightarrow C\%_{ZnSO_4}=\dfrac{16,1}{196+6,5}.100\%\approx7,95\%\)