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

Ta có: \(n_{H_2SO_4}=0,25.1=0,25\left(mol\right)\)

a, Theo PT: \(n_{NaOH}=2n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow V_{NaOH}=\dfrac{0,5}{2}=0,25\left(l\right)\)

b, Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)

\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,25}{0,25+0,25}=0,5\left(M\right)\)

nAl=10,8/27=0,4 mol

2Al +3H2SO4 --> Al2(SO4)3 + 3H2

0,4      0,6                                  0,6                 mol

=>VH2=0,6*22,4=13,44 lít

mH2SO4=0,6*98=58,8 g

C% H2SO4= 58,8*100/200=29,4 %

\(n_{CuO}=\dfrac{8}{80}=0.1\left(mol\right)\)

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

\(0.1...........0.1.........0.1\)

\(n_{NaOH}=0.24\cdot0.5=0.12\left(mol\right)\)

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

\(0.12..........0.06\)

\(n_{H_2SO_4}=0.1+0.06=0.16\left(mol\right)\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.16}{1}=0.16\left(l\right)\)

\(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0.06}{0.16}=0.375\left(M\right)\)

\(C_{M_{CuSO_4}}=\dfrac{0.1}{0.16}=0.625\left(M\right)\)

\(Đặt.oxit:A_2O_3\\ A_2O_3+3H_2SO_4\rightarrow A_2\left(SO_4\right)_3+3H_2O\\ n_{Al_2O_3}=\dfrac{34,2-10,2}{96.3-16.3}=0,1\left(mol\right)\\ M_{A_2O_3}=\dfrac{10,2}{0,1}=102\left(\dfrac{g}{mol}\right)=2M_A+48\\ \Rightarrow M_A=27\left(\dfrac{g}{mol}\right)\\ a,\Rightarrow A.là.nhôm\left(Al=27\right)\\ b,n_{H_2SO_4}=3.0,1=0,3\left(mol\right)\\ C\%_{ddH_2SO_4}=\dfrac{0,3.98}{100}.100=29,4\%\\ c,n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=0,1\left(mol\right)\\ Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3+3Na_2SO_4\\ n_{NaOH}=6.0,1=0,6\left(mol\right)\\ V_{ddNaOH}=\dfrac{0,6}{1,5}=0,4\left(l\right)\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

a, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)

b, \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{9,8}{9,8\%}=100\left(g\right)\)

c, Ta có: m _{dd sau pư} = 5,6 + 100 - 0,1.2 = 105,4 (g)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,1.152}{105,4}.100\%\approx14,42\%\)

\(n_{CaCO_3}=\dfrac{25}{100}=0.25\left(mol\right)\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

\(0.25...........0.25...........0.25\)

\(C_{M_{Ca\left(OH\right)_2}}=\dfrac{0.25}{0.1}=2.5\left(M\right)\)

\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

\(0.25..............0.25\)

\(V_{CH_4}=0.25\cdot22.4=5.6\left(l\right)\)

a) nCaCO3=0,25(mol)

CH4 + 2 O2 -to-> CO2 + 2 H2O

0,25<------------------0,25(mol)

CO2 + Ca(OH)2 -> CaCO3 + H2O

0,25<------0,25-----------0,25(mol)

b) CMddCa(OH)2= 0,25/0,1= 2,5(M)

b) V(CH4,đktc)=0,25.22,4=5,6(l)

a) Mg + H₂SO₄ --> MgSO₄ + H₂

b) \(n_{Mg}=\dfrac{3}{24}=0,125\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

         0,125->0,125-->0,125-->0,125

=> V_H2 = 0,125.22,4 = 2,8 (l)

\(V_{dd.H_2SO_4}=\dfrac{0,125}{2}=0,0625\left(l\right)\)

c) Sản phẩm là Magie sunfat và khí hidro

\(m_{MgSO_4}=0,125.120=15\left(g\right)\)

m_H2 = 0,125.2 = 0,25 (g)

d) 

\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,125}{1}\) => Hiệu suất tính theo H₂

Gọi số mol CuO bị khử là a (mol)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

               a--->a-------->a

=> 16 - 80a + 64a = 14,4 

=> a = 0,1 (mol)

=> n_H2(pư) = 0,1 (mol)

=> \(H=\dfrac{0,1}{0,125}.100\%=80\%\)