Goi $m_{dd\ H_2SO_4} = 139(gam)$
Ta có :

$V_{dd} = m : D = 139 : 1,39 = 100(ml)  = 0,1(lít)$

$n_{H_2SO_4} = 0,1.6,95 = 0,695(mol)$
Vậy : 

$C\%_{H_2SO_4} = \dfrac{0,695.98}{139}.100\% = 49\%$

\(C\%=\dfrac{C_M\cdot M_{H_2SO_4}}{10d}=\dfrac{6,95\cdot98}{13,9}=49\%\)

\(BTNT\left(S\right):n_{SO3}=n_{SO2}=\dfrac{V}{22,4}=\dfrac{5}{14}\left(mol\right)\)

Ta có : \(n_{H2SO4}=\dfrac{9}{25}\left(mol\right)\)

\(\Rightarrow\Sigma n_{H2SO4}=\dfrac{5}{14}+\dfrac{9}{25}=\dfrac{251}{350}\left(mol\right)\)

\(\Rightarrow C\%=\dfrac{m_{H2SO4}}{m_{dd}}.100\%=64,7\%\)

\(a) n_{Zn} = \dfrac{19,5}{65} = 0,3(mol) ; n_{HCl} = 0,35.2 = 0,7(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{HCl} = 0,7 > 2n_{Zn} = 0,6 \to HCl\ dư\\ n_{H_2} = n_{Zn} = 0,3(mol) \Rightarrow V_{H_2} = 0,3.22,4 = 6,72(lít)\\ b) m_{dd\ HCl} = 350.1,05 = 367,5(gam)\\ m_{dd\ sau\ pư} = 19,5 + 367,5 - 0,3.2 = 386,4(gam)\\ \Rightarrow C\%_{ZnCl_2} = \dfrac{0,3.136}{386,4}.100\% = 10,56\%\\ c) C\%_{HCl} = \dfrac{0,7.36,5}{367,5}.100\% = 6,95\%\)

$m_{dd} = 122,5 + 8 = 130,5(gam)$
$CuO + H_2SO_4 \to CuSO_4 + H_2O$

Theo PTHH :

$n_{CuSO_4} = n_{CuO} = 8 :80 = 0,1(mol)$
$C\%_{CuSO_4} = \dfrac{0,1.160}{130,5}.100\% = 12,26\%$

\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

PTHH: 2R + 3H₂SO₄ --> R₂(SO₄)₃ + 3H₂

____0,04<----0,06-------0,02<------0,06____(mol)
=> \(M_R=\dfrac{1,08}{0,04}=27\left(g/mol\right)=>Al\)

\(C_{M\left(H_2SO_4\right)}=\dfrac{0,06}{0,05}=1,2M\)

\(m_{ct}=\dfrac{19,6.200}{100}=39,2\left(g\right)\)

\(n_{H2SO4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

\(m_{ct}=\dfrac{5,2.200}{100}=10,4\left(g\right)\)

\(n_{BaCl2}=\dfrac{10,4}{208}=0,05\left(mol\right)\)

Pt : \(H_2SO_4+BaCl_2\rightarrow2HCl+BaSO_4|\)

            1             1             2             1

          0,4           0,05        0,1         0,05

a) Lập tỉ số so sánh: \(\dfrac{0,4}{1}>\dfrac{0,05}{1}\)

             ⇒ H₂SO₄ dư , BaCl₂ phản ứng hết

             ⇒ Tính toán dựa vào số mol của BaCl₂

\(n_{BaSO4}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

⇒ \(m_{BaSO4}=0,05.233=11,65\left(g\right)\)

b) \(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)

⇒ \(m_{HCl}=0,1.36,5=3,65\left(g\right)\)

\(n_{H2SO4\left(dư\right)}=0,4-0,05=0,35\left(mol\right)\)

⇒ \(m_{H2SO4\left(dư\right)}=0,35.98=34,3\left(g\right)\)

\(m_{ddspu}=200+200-11,65=388,35\left(g\right)\)

\(C_{ddHCl}=\dfrac{3,65.100}{388,35}=0,94\)⁰/₀

\(C_{ddH2SO4\left(dư\right)}=\dfrac{34,3.100}{388,35}=8,83\)⁰/₀

 Chúc bạn học tốt

\(a.n_{H_2SO_4}=\dfrac{200.19,6\%}{98}=0,4\left(mol\right)\\ n_{BaCl_2}=\dfrac{200.5,2\%}{208}=0,05\left(mol\right)\\ BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ Vì:\dfrac{0,05}{1}< \dfrac{0,4}{1}\\ \Rightarrow H_2SO_4dư\\ n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{BaSO_4}=0,05.233=11,65\left(g\right)\\ b.m_{ddsau}=200+200-11,65=388,35\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,05.2.36,5}{388,35}.100\approx0,94\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,05\right).98}{388,35}.100\approx8,832\%\)

Bài 1 :

Giả sử thể tích dung dịch H2SO4 là V ml

\(\rightarrow m_{dd}=1,84V\left(g\right)\rightarrow m_{H2SO4}=1,84V.98\%=1,8032\left(V\right)\)

\(\rightarrow n_{H2SO4}=\frac{1,8032V}{98}=0,0184V\left(mol\right)\)

\(\rightarrow CM_{H2SO4}=\frac{0,0184V.1000}{V}=18,4M\)

\(n_{H2SO4}=2.2,5=5\left(mol\right)\rightarrow m_{H2SO4}=5.98=490\left(g\right)\)

\(\rightarrow m_{dd_{H2SO4_{Can}}}=\frac{490}{98\%}=500\left(g\right)\)

Vậy V dung dịch H2SO4 cần \(=\frac{500}{1,84}=271,74\left(ml\right)\)

Cho 271,74 ml H2SO4 98% vào dung dịch, sau đó thêm H2O vào đủ 2 lít/

Bài 2:

Gọi số mol Na2O cần là x \(\rightarrow m_{Na2O}=62x\)

\(\rightarrow\) m dung dịch sau khi thêm=62x+84,5 gam

\(Na_2O+H_2O\rightarrow2NaOH\)

\(\rightarrow n_{NaOH_{tao.ra}}=2x\rightarrow m_{NaOH_{tao.ra}}=2x.40=80x\left(g\right)\)

\(\rightarrow\) m NaOH trong dung dịch \(=80x+84,5.10\%=80x+8,45\left(g\right)\)

\(\rightarrow C\%_{NaOH}=\frac{\left(80x+8,45\right)}{\left(62x+84,5\right)}=28,45\%\rightarrow x=0,25\)

\(\rightarrow m_{Na2O}=15,5\left(g\right)\)

Bài 3 :

\(n_{MgCO3}=\frac{16,8}{84}=0,2\left(mol\right)\)

\(n_{HCl}=\frac{200.10,95\%}{36,5}=0,6\left(mol\right)\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

Nên HCl dư

\(n_{CO2}=0,2\left(mol\right)\)

\(n_{HCl_{du}}=0,6-0,2.2=0,2\left(mol\right)\)

\(m_{dd_{Spu}}=16,8+200-0,2.44=208\left(g\right)\)

\(C\%_{HCl}=\frac{0,2.36,5}{208}.100\%=3,51\%\)

\(C\%_{MgCl2}=\frac{0,2.95}{208}.100\%=9,13\%\)