601,2=50(ml)nNaOH=20.20100.40=0,1(mol)" role="presentation" style="border:0px; box-sizing:inherit; direction:ltr; display:inline-block; float:none; font-size:18.08px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">601,2=50(ml)nNaOH=20.20100.40=0,1(mol)

0,050,05=1(M)" role="presentation" style="border:0px; box-sizing:inherit; direction:ltr; display:inline-block; float:none; font-size:18.08px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">(M)

Bài 9 : 

\(a) n_{Fe_2O_3} = \dfrac{3,2}{160}=0,02(mol)\\ Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O\\ n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,06(mol)\\ m_{dd\ H_2SO_4} = \dfrac{0,06.98}{19,6\%} = 30(gam)\\ b) \text{Chất tan : } Fe_2(SO_4)_3\\ n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = 0,02(mol)\\ m_{Fe_2(SO_4)_3} = 0,02.400 =8(gam)\)

Bài 8 : 

\(n_{Al_2O_3} = \dfrac{2,04}{102} = 0,02(mol)\\ Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O\\ n_{HCl} = 6n_{Al_2O_3} = 0,02.6 = 0,12(mol)\\ C_{M_{HCl}} = \dfrac{0,12}{0,2} =0,6M\)

\(n_{H_2SO_4}=\dfrac{200.9,8}{100.98}=0,2\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

______0,4<------0,2

=> \(V_{dd}=\dfrac{0,4}{1}=0,4\left(l\right)\)

\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02(mol)\\ PTHH:Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2\\ \Rightarrow n_{H_2SO_4}=3n_{Fe_2O_3}=0,06(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,06.98}{19,6\%}=30(g)\)

a) \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

\(HCl+NaOH\rightarrow NaCl+H_2O\)

b) \(n_{CaO}=\dfrac{2,8}{56}=0,05\left(mol\right);n_{NaOH}=\dfrac{100.4\%}{40}=0,1\left(mol\right)\)

\(m_{muối}=m_{CaCl_2}+m_{NaCl}=0,05.111+0,1.58,5=11,4\left(g\right)\)

c) \(CM_{HCl}=\dfrac{0,05.2+0,1}{0,5}=0,4M\)

a, Ta có: \(m_{NaOH}=200.4\%=8\left(g\right)\) \(\Rightarrow n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

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

_____0,2______0,1_______0,1 (mol)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,1}{0,2}=0,5M\)

b, Ta có: m _{dd sau pư} = m _{dd NaOH} + m _ddH2SO4 = 200 + 50 = 250 (g)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,1.142}{250}.100\%=5,68\%\)

Bạn tham khảo nhé!

a)

$CuO + H_2SO_4 \to CuSO_4 + H_2O$

$n_{H_2SO_4} = n_{CuO} = \dfrac{1,6}{80} = 0,02(mol)$
$C\%_{H_2SO_4} = \dfrac{0,02.98}{100}.100\% = 1,96\%$

b)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4} = 0,04(mol)$
$m_{NaOH} = 0,04.40 = 1,6(gam)$

c)

$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$

Cu dư nên $n_{SO_2} = \dfrac{1}{2}n_{H_2SO_4} = 0,05(mol)$
$V_{SO_2} = 0,05.22,4 = 1,12(lít)$

\(n_{Fe_2O_3}=\dfrac{48}{160}=0,3(mol)\\ PTHH:Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\ \Rightarrow n_{H_2SO_4}=3n_{Fe_2O_3}=0,9(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,9.98}{19,6\%}=450(g)\)