Question

Cho 24g đồng ( II) oxit tác dụng với 100g đ H2SO4 24.5%. Tính C% các chất có trong dd sau khi phản ứng kết thúc

Answer 1

PTHH: CuO + \(H_2SO_4->CuSO_4+H_2O\)

Ta có : \(n_{CuO}=\frac{20}{80}=0,25mol\)

n\(_{H_2SO_4}\)=\(\frac{25,5.100}{98.100}=0,26mol\)

=> \(H_2SO_4dư\)

Theo pthh: \(n_{H_2SO_4}=n_{CuO}=0,25mol\)

\(n_{H_2SO_4dư}=0,26-0,25=0,1mol\)

C%\(_{H_2SO_4}=\frac{0,1.9,8}{100}.100\%=9,8\%\)

Theo pthh : \(n_{CúSO_4}=n_{CuO}=0,25mol\)

C%\(_{C_{ }uSO_4}=\frac{0,25.160}{100}.100\%=40\%\)

Answer 2

PTHH: CuO + H₂SO₄ ---> CuSO₄ + H₂O

Mol: 0,25 0,25 0,25

n_CuO = 24/80 = 0,3 mol

m_H2SO4 = 24,5% *100/ 100% = 24,5g

n_H2SO4 = 24,5/98 = 0,25 mol

Ta có tỉ lệ: n_CuO = 0,3 > n_H2SO4 = 0,25

=> CuO dư, H₂SO₄ hết

n_{CuO dư} = 0,3 - 0,25 = 0,05 mol

m_{CuO dư} = 24 - 0,05*80 = 20g

m_CuSO4 = 0,25*160 = 40g

m_{dd sau pư} = 100 + 24 = 124g

C%_{CuO dư} = 20*100% / 124 = 16,13%

C%_CuSO4 = 40*100% / 124 = 32,3%

Answer 3

\(m_{H_2SO_4}=\frac{100.24,5}{100}=24,5\left(g\right)\\ \rightarrow n_{H_2SO_4}=\frac{24,5}{98}=0,25\left(mol\right)\\ n_{Cu}=\frac{24}{64}=0,375\left(mol\right)\)

\(PTHH:Cu+2H_2SO_4\underrightarrow{t^o}CuSO_4+2H_2O+SO_2\)

\(TL:\frac{0,375}{1}>\frac{0,25}{2}\rightarrow Cu.du\)

\(m_{Cu.du}=64.\left(0,375-0,125\right)=16\left(g\right)\)

\(m_{CuSO_4}=0,125.160=20\left(g\right)\)

\(m_{SO_2}=0,125.64=8\left(g\right)\)

\(m_{ddspu}=24+100-8=116\left(g\right)\)

\(C\%_{Cu.du}=\frac{16}{116}.100\%=13,79\left(\%\right)\)

\(C\%_{CuSO_4}=\frac{20}{116}.100\%=17,24\left(\%\right)\)

Answer 4

Chưa đọc kĩ đề :)

\(n_{H_2SO_4}=\frac{100.24,5}{100.98}=0,25\left(mol\right)\\ n_{CuO}=\frac{24}{80}=0,3\left(mol\right)\\ PTHH:CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ TL:\frac{0,25}{1}< \frac{0,3}{1}\rightarrow CuO.du\\ m_{ddspu}=24+100=124\left(g\right)\\ C\%_{CuSO_4}=\frac{0,25.160}{124}.100\%=32,36\left(\%\right)\\ C\%_{CuO.du}=\frac{80.\left(0,3-0,25\right)}{124}.100\%=3,26\left(\%\right)\)

Answer 5

https://i.imgur.com/F4kTva8.jpg

Answer 6

nCuO = 0.3 mol

mH2SO4 = 24.5 g

nH2SO4 = 0.5 mol

CuO + H2SO4 --> CuSO4 + H2O

Bđ: 0.3____0.5

Pư: 0.3____0.3________0.3

Kt: 0______0.2________0.3

mH2SO4 dư = 19.6 g

mCuSO4 = 48 g

mdd sau phản ứng = 24 + 100 = 124 (g)

C%H2SO4 = 15.8%

C%CuSO4 = 38.71%

Answer 7

nCuO = 0.3 mol

mH2SO4 = 24.5 g

nH2SO4 = 0.25 mol

CuO + H2SO4 --> CuSO4 + H2O

Bđ: 0.3____0.25

Pư: 0.25____0.25___0.25

Kt: 0.05____0_____0.25

mCu dư = 4 g

mCuSO4 = 40 g

mdd sau phản ứng = 24 + 100 - 4 = 120 (g)

C%CuSO4 = 33.33%