a)

$CuO + H_2SO_4 \to CuSO_4 + H_2O$
$n_{CuO} = \dfrac{8}{80} = 0,1 < n_{H_2SO_4} = 0,2.1 = 0,2$ nên $H_2SO_4$ dư

Theo PTHH : $n_{CuSO_4} = n_{CuO} = 0,1(mol)$
$m_{CuSO_4} = 0,1.160 = 16(gam)$

b)

$n_{H_2SO_4\ dư} = 0,2  - 0,1 = 0,1(mol)$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,1}{0,2} = 0,5M$
$C_{M_{CuSO_4}} = \dfrac{0,1}{0,2} = 0,5M$

nFe3O4= 23,2/232=0,1(mol)

nH2SO4=2.0,25=0,5(mol)

a) PTHH: Fe3O4 + 4 H2SO4 -> Fe2(SO4)3 + FeSO4 + 4 H2O

Ta có: 0,1/1 < 0,5/4

=> H2SO4 dư, Fe3O4 hết, tính theo Fe3O4

nFe2(SO4)3=nFeSO4=nFe3O4=0,1(mol)

=> mFe2(SO4)3=0,1.400=40(g)

mFeSO4=0,1.152=15,2(g)

b) nH2SO4(dư) = 0,5- 0,1.4=0,1(mol)

Vddsau=VddH2SO4=0,25(l)

=> CMddH2SO4(dư)=CMddFe2(SO4)3=CMddFeSO4=0,1/0,25=0,4(M)

a, PT: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

Gọi: \(\left\{{}\begin{matrix}n_{CuO}=x\left(mol\right)\\n_{FeO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 80x + 72y = 11,2 (1)

Ta có: \(n_{H_2SO_4}=0,15.1=0,15\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{CuO}+n_{FeO}=x+y=0,15\left(2\right)\)

Từ (1) và (2) ⇒ x = 0,05 (mol), y = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,05.80}{11,2}.100\%\approx35,71\%\\\%m_{FeO}\approx64,28\%\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{CuSO_4}=n_{Cu}=0,05\left(mol\right)\\n_{FeSO_4}=n_{FeO}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{CuSO_4}}=\dfrac{0,05}{0,15}=\dfrac{1}{3}\left(M\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,15}=\dfrac{2}{3}\left(M\right)\end{matrix}\right.\)

PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(n_{Na_2SO_4}=n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot\dfrac{10}{40}=0,125\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,125\cdot98}{10\%}=122,5\left(g\right)\\m_{Na_2SO_4}=0,125\cdot142=17,75\left(g\right)\end{matrix}\right.\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{17,75}{10+122,5}\cdot100\%\approx13,4\%\)

300ml = 0,3l

\(n_{HNO3}=1.0,3=0,3\left(mol\right)\)

Pt : \(NaOH+HNO_3\rightarrow NaNO_3+H_2O|\)

          1               1               1              1

         0,3            0,3            0,3

\(n_{NaOH}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

200ml = 0,2l

\(C_{M_{NaOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

\(n_{NaNO3}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

⇒ \(m_{NaNO3}=0,3.85=25,5\left(g\right)\)

Sau phản ứng :

\(V_{dd}=0,2+0,3=0,5\left(l\right)\)

\(C_{M_{NaNO3}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)

 Chúc bạn học tốt

\(n_{HNO_3}=0,3\left(mol\right)\)

\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

Theo PT: \(n_{NaOH}=n_{NaNO_3}=n_{HNO_3}=0,3\left(mol\right)\)

\(\Rightarrow CM_{NaOH}=\dfrac{0,3}{0,2}=1,5M\)

\(m_{NaNO_3}=0,3.85=25,5\left(g\right)\)

Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{6}\left(mol\right)\Rightarrow m_{Al}=\dfrac{1}{6}.27=4,5\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=0,25\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)

c, \(n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{12}\left(mol\right)\)

\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=\dfrac{1}{12}.342=28,5\left(g\right)\)

a) \(n_{NaOH}=0,2.1=0,2\left(mol\right);n_{H_2SO_4}=0,15.2=0,3\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:         0,2                            0,1

Ta có: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) ⇒ NaOH hết, H₂SO₄ dư

\(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

b) V_{dd sau pứ} = 0,2 + 0,15 = 0,35 (l)

\(C_{M_{ddNa_2SO_4}}=\dfrac{0,1}{0,35}=\dfrac{2}{7}\approx0,2857M\)

\(C_{M_{ddH_2SO_4dư}}=\dfrac{0,3-0,1}{0,35}=\dfrac{4}{7}\approx0,57M\)

n H2SO4=\(\dfrac{10\%.490}{2+32+16.4}=0,5mol\)

n Al2O3 =\(\dfrac{10,2}{27.2+16.3}=0,1mol\)

       \(Al_2O_3+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2O\)

bđ    0,1............0,5

pư    0,1............0,3..................0,1

spu   0 ................0,2................0,1 

=> sau pư gồm H2SO4 dư , Al2(S04)3 và H2O

m H2SO4 dư = \(0,2.\left(2+32+16.3\right)=19,6g\)

m Al2(SO4)3 = \(0,1\left(27.2+32.3+16.4.3\right)=34,2g\)

m dd = \(490+10,2=500,2g\)

% Al2(SO4)3 = \(\dfrac{34,2}{500,2}.100\sim6,84\%\)

% H2SO4 dư = \(\dfrac{19,6}{500,2}.100\sim3,92\%\)

a, \(n_{KOH}=0,3.1=0,3\left(mol\right)\)

PT: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

Theo PT: \(n_{K_2SO_4}=\dfrac{1}{2}n_{KOH}=0,15\left(mol\right)\)

\(\Rightarrow m_{K_2SO_4}=0,15.174=26,1\left(g\right)\)

b, \(C_{M_{K_2SO_4}}=\dfrac{0,3}{0,3+0,2}=0,6\left(M\right)\)