Question

Cho 200ml dd CuSO4 1M tac dung vua du voi 50ml đe NaOH thu ket tua loc lay ket tua nung nong trg khong khi o nhiet do cao den khoi luong ko doi thu m gam chat ran .

A) tinh nong do mol đe NAOH

B) tìm m

Answer 1

CuSO4 + 2NaOH ---> Cu(OH)2 + Na2SO4

0.2 0.4 0.2

Cu(OH)2----> CuO+ H2O

0.2 0.2

nCuSO4= 1.0,2=0,2mol

CM NaOH= 0,4/02=2M

mCuo= 0,2x80=16(g)

Answer 2

CuSO₄ + 2NaOH → Na₂SO₄ + Cu(OH)₂↓ (1)

Cu(OH)₂ \(\underrightarrow{to}\) CuO + H₂O (2)

\(n_{CuSO_4}=0,2\times1=0,2\left(mol\right)\)

a) Theo PT1: \(n_{NaOH}=2n_{CuSO_4}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,4}{0,05}=8\left(M\right)\)

b) Theo Pt1: \(n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,2\left(mol\right)\)

Theo PT2: \(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,2\times80=16\left(g\right)\)

Vậy \(m=16\left(g\right)\)

Answer 3

2NaOH + CuSO₄ → Na₂SO₄ + Cu(OH)₂↓ (1)

Cu(OH)₂ \(\underrightarrow{to}\) CuO + H₂O (2)

\(n_{CuSO_4}=0,2\times1=0,2\left(mol\right)\)

a) Theo PT1: \(n_{NaOH}=2n_{CuSO_4}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,4}{0,05}=8\left(M\right)\)

b) Theo Pt1: \(n_{Cu\left(OH\right)_2}=n_{CuSO_4}=0,2\left(mol\right)\)

Theo PT2: \(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,2\times80=16\left(g\right)\)

Vậy \(m=16\left(g\right)\)

Answer 4

a/ CuSO4 + 2NaOH ==> Cu(OH)2 + Na2SO4

0,2 ................0,4..................0,2

Cu(OH)2 ==> CuO + H2O

0,2...................0,2

[NaOH] = 0,4 : 0,05 = 8M

b/ mCuO = 0,2 x 80 = 16g