CuO+H2SO4--->CuSO4+H2O

b) n CuO=1,6/80=0,02(mol)

m H2SO4=100.20/100=20(g)

n H2SO4=20/98=0,2(mol)

--->H2SO4 dư

dd sau pư là H2SO4 dư và CuSO4

m dd sau pư=1,6+100=101,6(g)

n H2SO4=n CuO=0,02(mol)

m H2SO4=0,02.98=1,96(g)

m H2SO4 dư=20-1,96=18,04(g)

C% H2SO4 dư=18,04/101,6.100%=17,76%

n CuSO4=n CuO=0,02(mol)

m CuSO4=0,02.160=3,2(g)

C% CuSO4=3,2/101,6.100%=3,15%

1, n SO2=0,4 mol

pthh

SO2 + 2H2O+Br2 ---> H2SO4+2HBr

0,4.......0,8.......0,4.........0,4...........0,8

chỉ có thể tính được Cm thôi

CM H2SO4=0,4:2=0,2 M

CM HBr=0,8:2=0,4 M

\(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

\(m_{H_2SO_4}=100.20\%=20\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{20}{98}=\dfrac{10}{49}\left(mol\right)\)

PTHH: CuO + H₂SO₄ → CuSO₄ + H₂O

Mol:      0,02      0,02          0,02

Ta có: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) ⇒ CuO hết, H₂SO₄ dư

m_{dd sau pứ} = 1,6 + 100 = 101,6 (g)

\(C\%_{ddCuSO_4}=\dfrac{0,02.160.100\%}{101,6}=3,15\%\)

\(C\%_{ddH_2SO_4}=\dfrac{\left(\dfrac{10}{49}-0,02\right).98.100\%}{101,6}=17,76\%\)

\(n_{CuO}=\dfrac{1,6}{80}=0,02mol\\ n_{H_2SO_4}=\dfrac{100.20}{100.98}=\dfrac{10}{49}mol\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ \Rightarrow\dfrac{0,02}{1}< \dfrac{10:49}{1}\Rightarrow H_2SO_4.dư\\ n_{CuO}=n_{CuSO_4}=n_{H_2SO_4,pư}=0,02mol\\ C_{\%CuSO_4}=\dfrac{0,02.160}{1,6+100}\cdot100=3,15\%\\ C_{\%H_2SO_4}=\dfrac{\left(10:49-0,02\right)98}{1,6+100}\cdot100=17,76\%\%\)

Ta có: \(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

\(m_{H_2SO_4}=100.20\%=20\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{20}{98}=\dfrac{10}{49}\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{CuSO_4}=n_{H_2SO_4\left(pư\right)}=n_{CuO}=0,02\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=\dfrac{10}{49}-0,02=\dfrac{451}{2450}\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{CuSO_4}=\dfrac{0,02.160}{1,6+100}.100\%\approx3,15\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{\dfrac{451}{2450}.98}{1,6+100}.100\%\approx17,76\%\end{matrix}\right.\)

https://hoc24.vn/hoi-dap/question/280059.html Câu tương tự.

Ahjhj! Hôm nay mk ms đi hx bài đâu tiên của lớp 9 về!

a) PTHH: CuO + H₂SO₄ --> CuSO₄ + H₂O

b) Ta có: n_CuO = \(\dfrac{1,6}{80}\) = 0,02 mol

\(m_{H_2SO_4}\) = 100.20% = 20 g => \(n_{H_2SO_4}\) = \(\dfrac{20}{98}\) = \(\dfrac{10}{49}\) mol

Vì \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) => H₂SO₄ dư

Áp dụng ĐLBTKL ta có:

m_CuO + m_H2SO4 = m_{dd sau p/ứ} = 1,6 + 100 = 101,6 g

Cứ 1 mol CuO --> 1 mol H₂SO₄ --> 1 mol CuSO₄

0,02 mol --> 0,02 mol --> 0,02 mol

=> C% của CuSO₄ = \(\dfrac{0,02.160}{101,6}.100\%\) = 3,15%

C% của H₂SO₄ dư = \(\dfrac{\left(\dfrac{10}{49}-0,02\right).98}{101,6}.100\%\) = 17,76%

\(\text{a)2Na+2H2O}\rightarrow\text{2NaOH+H2}\)

\(\text{2Al+2NaOH+2H2O}\rightarrow\text{2NaAlO2+3H2}\)

Chất rắn không tan là Al dư

\(\text{2Al+3H2SO4}\rightarrow\text{Al2(SO4)3+3H2}\)

\(\text{2H2O+NaAlO2+CO2}\rightarrow\text{Al(OH)3+NaHCO3}\)

\(\text{Al2(SO4)3+6H2O+6NH3}\rightarrow\text{3(NH4)2SO4+2Al(OH)3}\)

\(\text{2Al(OH)3}\rightarrow\text{Al2O3+3H2O}\)

b) nAl dư =2/3xnH2=2/3x0,15=0,1(mol)

gọi a là số mol Na

Ta có:

\(\text{a/2+3a/2=0,4=>a=0,2(mol)}\)

\(\left\{{}\begin{matrix}\text{mNa=0,2x23=4,6(g)}\\\text{mAl=(0,2+0,1)x27=8,1(g)}\end{matrix}\right.\)