Question

hoà tan hết 1,6 gam Cu trong dung dịch axit sunfuric đặc(H2SO4 đặc 98 ),đun nóng thu được V ml khí SO2 (ở ĐKTC)

Tính V và khối lượng dung dịch H2SO4 đặc đã tham gia phản ứng

Answer 1

PTHH: Cu+ H2SO4 ---> CUO + SO2+ H2O
tính nCu=1,6:64=0,025 mol
Theo PTHH có nH2O=nCu=0,025 mol
Suy ra VSO2=0,025*22,4=0,56

Answer 2

PTHH: \(Cu+2H_2SO_4\rightarrow CuSO_4+SO_2+2H_2O\)
\(n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\)
Theo PT ta có: \(n_{Cu}=n_{SO_2}=0,025\left(mol\right)\)
\(\Rightarrow V_{SO_2}=0,025.22,4=0,56\left(l\right)\)
Theo PT ta có: \(n_{H_2SO_4}=2.n_{Cu}=2.0,025=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)

Answer 3

PTHH: \(Cu+2H_2SO_4\rightarrow2H_2O+SO_2+CuSO_4\\ 0,025mol:0,05mol\rightarrow0,05mol:0,025mol\)

\(n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\)

\(V_{SO_2}=0,025.22,4=0,56\left(l\right)\)

\(m_{H_2SO_4}=0,05.98=4,9\left(g\right)\)

Answer 4

n_Cu = \(\dfrac{m}{M}=\dfrac{1,6}{64}=0,025\left(mol\right)\)

PTHH : Cu + 2H₂SO₄ ---> CuSO₄ + 2H₂O + SO₂

0,025 0,05 0,025 0,05 0,025 ( mol)

=> \(V_{SO_2}=n\cdot22,4=0,025\cdot22,4=0,56\left(l\right)\)

=> \(m_{H_2SO_4}=n\cdot M=0,05\cdot98=4,9\left(g\right)\)

ta có C%= 98%

=> \(mdd_{H_2SO_4}=\dfrac{m_{ct}\cdot100}{C\%}=\dfrac{4,9\cdot100}{98}=5\left(g\right)\)

hok tốt nhé

Answer 5

\(n_{Cu}=\dfrac{1,6}{64}=0,025\left(mol\right)\)

Theo PTHH: \(Cu+2H_2SO_4\rightarrow CuSO_4+SO_2+2H_2O\)

Theo PTHH: \(n_{Cu}=n_{SO_2}=1:1\)

\(\Rightarrow n_{Cu}=n_{SO_2}=0,025\left(mol\right)\)

\(\Rightarrow V_{SO_2\left(đktc\right)}=n_{SO_2}.22,4=0,025.22,4=0,56\left(l\right)\)

PTHH: \(Cu+2H_2SO_4\rightarrow CuSO_4+SO_2+H_2O\)

1mol → 2 mol

0,025mol→ a mol

Gọi a là số mol của \(H_2SO_4\)

Theo PTHH ta có: \(n_{H_2SO_4}=a=\dfrac{0,025\times2}{1}=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=n_{H_2SO_4}.M_{H_2SO_4}=0,05.98=4,9\left(g\right)\)

Answer 6

Có n\(cu\)=\(\dfrac{m}{M}\) =\(\dfrac{1.6}{64}\)=0.025(mol)

PT: Cu +2H₂SO₄ →Cu SO₄+SO_2^↑+2H₂O

(mol):0,025→0,05→0,025→0,025→0,05

a) V_SO2=n.22.4=0,025.22,4=0.56(l)

b)m_H2SO4=0.05.98=4.9(g)

⇒mdd_{H2SO4 98% đã dug=\(\dfrac{mct}{C\%}\) =\(\dfrac{4.9}{98\%}\) =5(g)}

Answer 7

n_Cu = mM=1,664=0,025(mol)mM=1,664=0,025(mol)

PTHH : Cu + 2H₂SO₄ ---> CuSO₄ + 2H₂O + SO₂

0,025 0,05 0,025 0,05 0,025 ( mol)

=> VSO2=n⋅22,4=0,025⋅22,4=0,56(l)VSO2=n⋅22,4=0,025⋅22,4=0,56(l)

=> mH2SO4=n⋅M=0,05⋅98=4,9(g)mH2SO4=n⋅M=0,05⋅98=4,9(g)

ta có C%= 98%

=> mddH2SO4=mct⋅100C%=4,9⋅10098=5(g)

Answer 8

 khi lần lượt cho các chất: CO², CuO, FE²O³,Mg, Ba(OH)² tác dụng với dung dịch H²SO⁴ loãng