\(n_{CuO}=\dfrac{3,2}{80}=0,04\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{100.20\%}{98}=0,204\left(mol\right)\)

PTHH:

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

0,04      0,04           0,04    

\(\dfrac{0,04}{1}< \dfrac{0,204}{1}\) --> H2SO4 dư

\(C\%_{CuSO_4}=\dfrac{0,04.160}{3,2+100}.100\%=6,2\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{3,2+100}.100\%=19\%\)

MgO+H2SO4->MgSO4+H2O

0,1-----0,1-------0,1--------0,1 mol

n MgO=4\40=0,1 mol

=>m H2SO4=0,1.98=9,8g

C%H2SO4=9,8\100 .100=9,8%

m MgSO4=0,1.120=12g

m muối =4+100-(0,1.18)=102,2g

=>C% muối=12\102,2.100=11,74 %

Câu 1:

CuO + H₂SO₄ → CuSO₄ + H₂O

\(n_{CuO}=\frac{3,2}{80}=0,04\left(mol\right)\)

\(m_{H_2SO_4}=200\times9,8\%=19,6\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\frac{19,6}{98}=0,2\left(mol\right)\)

Theo PT: \(n_{CuO}=n_{H_2SO_4}\)

Theo bài: \(n_{CuO}=\frac{1}{5}n_{H_2SO_4}\)

Vì \(\frac{1}{5}< 1\) ⇒ H₂SO₄ dư

Dung dịch sau pư gồm: H₂SO₄ dư và CuSO₄

Ta có: \(m_{dd}saupư=3,2+200=203,2\left(g\right)\)

Theo Pt: \(n_{H_2SO_4}pư=n_{CuO}=0,04\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}dư=0,2-0,04=0,16\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}dư=0,16\times98=15,68\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}dư=\frac{15,68}{203,2}\times100\%=7,72\%\)

Theo Pt: \(n_{CuSO_4}=n_{CuO}=0,04\left(mol\right)\)

\(\Rightarrow m_{CuSO_4}=0,04\times160=6,4\left(g\right)\)

\(\Rightarrow C\%_{CuSO_4}=\frac{6,4}{203,2}\times100\%=3,15\%\)

Câu 2:

ZnO + H₂SO₄ → ZnSO₄ + H₂O

\(n_{ZnO}=\frac{8,1}{81}=0,1\left(mol\right)\)

\(m_{H_2SO_4}=200\times24,5\%=49\left(g\right)\)

\(\Rightarrow n_{H_2SO_4}=\frac{49}{98}=0,5\left(mol\right)\)

Theo Pt: \(n_{ZnO}=n_{H_2SO_4}\)

Theo bài: \(n_{ZnO}=\frac{1}{5}n_{H_2SO_4}\)

Vì \(\frac{1}{5}< 1\) ⇒ H₂SO₄ dư

Dung dịch sau pư gồm: H₂SO₄ dư và ZnSO₄

Ta có: \(m_{dd}saupư=8,1+200=208,1\left(g\right)\)

Theo PT: \(n_{H_2SO_4}pư=n_{ZnO}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}dư=0,5-0,1=0,4\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,4\times98=39,2\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\frac{39,2}{208,1}\times100\%=18,84\%\)

Theo pT: \(n_{ZnSO_4}=n_{ZnO}=0,1\left(mol\right)\)

\(\Rightarrow m_{ZnSO_4}=0,1\times161=16,1\left(g\right)\)

\(\Rightarrow C\%_{ZnSO_4}=\frac{16,1}{208,1}\times100\%=7,74\%\)

a)m_H2SO4=\(\dfrac{200.7,3\text{%}}{100\%}\)=14,6g

n_HCl=\(\dfrac{14,6}{36,5}\)=0,4(mol)

PTHH:

NaOH+ HCl→ NaCl+ H₂O

1            1            1         1

0,4        0,4         0,4     (mol)

⇒m_NaOH=0,4.40=16(g)

Nồng độ % của dd NaOH cần dùng là:

C_%NaOH=\(\dfrac{16}{200}\)   .100%=8%

b)Ta có:m_{dd spứ}=m_{dd trc pứ}=400g

m_NaCl=0,4.58,5=23,4g

Nồng độ % dd muối tạo thành sau pứ là:

C_%dd NaCl=\(\dfrac{23,4}{400}\)   .100%=5,85%        

ncuo= 1,6/80=0,02

nh2so4=(100*20)/( 98*100)= 0,2> 0,02-> cuo pư hết, h2so4 dư

cuo+ h2so4-> cuso4+h2o

0,02-> 0,02 0,02

mdd sau pư= 1,6+ 100= 101,6

c%h2so4 dư= (0,2-0,02)*98/101,6*100= 17,36%

c%cuso4= 0,02*160/101,6*100= 3,15%

n_CuO= \(\frac{1,6}{80}\) = 0,02 (mol)

\(n_{H_2SO_4}\) = \(\frac{100.20\%}{98}\) =0,2041(mol)

CuO + H₂SO₄ \(\rightarrow\) CuSO₄ + H₂O

bđ 0,02 \(\frac{10}{49}\) (mol)

pư 0,02 \(\rightarrow\) 0,02 \(\rightarrow\) 0,02 (mol)

spư 0 0,1841 0,02 (mol)

m_{d d} (sau pư) = 100 + 1,6 = 101,6 (g)

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

C%(H₂SO₄)= \(\frac{0,1841.98}{101,6}\) . 100% = 17,76%

pthh

CuO+ H2SO4--------> CuSO4 + H2O

0,02 0,02 0,02 0,02

m H2SO4 dư = 20-19,6=0,4 g

m CuSO4= 0,02.160=3,2 g

m dung dịch =1,6 + 100= 101,6 g

C% H2SO4=0,39 %

C% CuSO4=3,15 %