2

b

mNaCl=\(\dfrac{200.15}{100}\)=30(g)

nNaCl=\(\dfrac{30}{58,5}\)=0.51(mol)

VddNaCl=\(\dfrac{200}{1,1}\)=181.8(ml)=0.1818(l)

CMNaCl=\(\dfrac{0,51}{0,1818}\)=2.8(M)

a.Mg + H2SO4 -> MgSO4 + H2

b.\(nH2=\dfrac{4.704}{22.4}=0.21mol\) = nMg

mMg = 0.21\(\times24=5.04g\)

\(\%mMg=\dfrac{5.04\times100}{25}=20.16\%\)

\(\%mAg=100-20.16=79.84\%\)

c.MgSO4 + 2KOH -> K2SO4 + Mg(OH)2

   0.21           0.42

H2SO4 + 2KOH -> K2SO4 + H2O

  0.04        0.08

\(nH2SO4=\dfrac{9.8\times250}{100\times98}=0.25mol\)

Mà nH2SO4 phản ứng = nH2 = 0.21 mol 

\(\Rightarrow nH2SO4dư=0.25-0.21=0.04mol\)

=> nKOH = 0.42 + 0.08 = 0.5mol

\(\Rightarrow CM_{KOH}=\dfrac{0.5}{0.625}=0.8M\)

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(m_{H_2SO_4}=100.9,8\%=9,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

PTHH: CuO + H₂SO₄ → CuSO₄ + H₂O

Mol:      0,05      0,05          0,05

Ta có: \(\dfrac{0,05}{1}< \dfrac{0,1}{1}\) ⇒ CuO hết, H₂SO₄ dư

\(C\%_{ddCuSO_4}=\dfrac{0,05.160.100\%}{4+100}=7,69\%\)

\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,1-0,05\right).98.100\%}{4+100}=4,71\%\)

\(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(m_{ct}=\dfrac{9,8.100}{100}=9,8\left(g\right)\)

\(n_{H2SO4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)

         1            1                1             1

       0,05       0,1             0,05

Lập tỉ số so sánh : \(\dfrac{0,05}{1}< \dfrac{0,1}{1}\)

               ⇒ CuO phản ừng hết , H₂SO₄ dư

               ⇒ Tính toán dựa vào số mol của CuO

\(n_{CuSO4}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

⇒ \(m_{CuSO4}=0,05.160=8\left(g\right)\)

\(n_{H2SO4\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)

⇒ \(m_{H2SO4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

\(m_{ddspu}=4+100=104\left(g\right)\)

\(C_{CuSO4}=\dfrac{8.100}{104}=7,69\)⁰/₀

\(C_{H2SO4\left(dư\right)}=\dfrac{4,9.100}{104}=4,71\)⁰/₀

 Chúc bạn học tốt

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

Pt : \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O|\)

          1           1                1            1

       0,02                         0,02

\(n_{CuSO4}=\dfrac{0,02.1}{1}=0,02\left(mol\right)\)

⇒ \(m_{CuSO4}=0,02,160=3,2\left(g\right)\)

\(m_{ddspu}=1,6+300=301,6\left(g\right)\)

\(C_{CuSO4}=\dfrac{3,2.100}{301,6}=1,6\)⁰/₀

 Chúc bạn học tốt

a) Ta có: \(m_{H_2SO_4}=120\cdot10\%=12\left(g\right)\) \(\Rightarrow C\%_{H_2SO_4}=\dfrac{12}{120+30}\cdot100\%=8\%\)

b) Ta có: \(m_{KOH}=195\cdot8\%=15,6\left(g\right)\) \(\Rightarrow C\%_{KOH}=\dfrac{15,6+5}{195+5}\cdot100\%=10,3\%\)

May quá bảo hỏi hộ đỡ mất công viết

Đáp án D

KOH + HCl → KCl + H₂O

TH1: n_HCl ≥ n_KOH ⇒ Chất tan chỉ gồm KCl

 n_KCl = n_KOH = 0,1

⇒ m_KCl = 7,45 > m _{chất tan} ⇒ loại

TH2: n_HCl < n_KOH

Đặt n_HCl = a , n_{KOH dư} = b ⇒ n _KOH = n _HCl + n _{KOH dư} = a + b = 0,1 mol

n _KCl = a mol

m _{chất tan} = m _KCl + m _{KOH dư} = 74,5a  + 56b = 6,525g

⇒ a = 0,05 mol; b = 0,05 mol

⇒ C_{M HCl} = 0,5

\(n_{CuSO_4}=\dfrac{50}{250}=0.2\left(mol\right)\)

\(n_{FeSO_4}=\dfrac{27.8}{278}=0.1\left(mol\right)\)

\(C_{M_{CuSO_4}}=C_{M_{FeSO_4}}=\dfrac{0.1}{0.1964}=0.5\left(M\right)\)

\(m_{dd_A}=50+27.8+196.4=274.2\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.1\cdot160}{274.2}\cdot100\%=6.47\%\)

\(C\%_{FeSO_4}=\dfrac{0.1\cdot152}{274.2}\cdot100\%=5.54\%\)

\(n_{CuSO_4.5H_2O}=\dfrac{50}{250}=0,2\left(mol\right)\)

=> \(m_{CuSO_4}=0,2.160=32\left(g\right)\)

\(m_{H_2O}=0,2.5.18=18\left(g\right)\)

=> \(m_{FeSO_4}=0,1.152=15,2\left(g\right)\)

\(m_{H_2O}=0,1.7.18=12,6\left(g\right)\)

\(m_{dd}=196,4+50+27,8=274,2\left(g\right)\)

\(V_{dd}=\dfrac{196,4+18+12,6}{1000}=0,227\left(l\right)\)

=> \(CM_{CuSO_4}=\dfrac{0,2}{0,227}=0,72M\)

\(C\%_{CuSO_4}=\dfrac{32}{274,2}.100=11,67\%\)

\(CM_{FeSO_4}=\dfrac{0,1}{0,227}=0,44M\)

\(C\%_{CuSO_4}=\dfrac{15,2}{274,2}.100=5,54\%\)

Này không cho D hả em?