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$n_{Na_2CO_3} = \dfrac{265.10\%}{106} = 0,25(mol)$
$n_{CaCl_2} = \dfrac{475,72.7\%}{111} = 0,3(mol)$
$CaCl_2 + Na_2CO_3 \to CaCO_3 + 2NaCl$

Ta thấy : $n_{CaCl_2} > n_{Na_2CO_3}$ nên $CaCl_2$ dư

$n_{CaCl_2\ dư} = 0,3 - 0,25 = 0,05(mol)$
$n_{NaCl} = 0,5(mol)$
Sau phản ứng, $m_{dd} = 265 + 475,72 - 0,25.100 = 715,72(gam)$
$C\%_{NaCl} = \dfrac{0,5.58,5}{715,72}.100\% = 4,09\%$
$C\%_{CaCl_2\ dư} = \dfrac{0,05.111}{715,72}.100\% = 0,76\%$

\(m_{Na_2CO_3}\) = \(265\times10\%\) = \(26,5\left(g\right)\) \(\Rightarrow n_{Na_2CO_3}\) = 0,25

\(m_{CaCl_2}\) = \(500\times6,6\%\) = \(33\left(g\right)\) \(\Rightarrow n_{CaCl_2}\)= \(\dfrac{11}{37}\)

 \(Na_2CO_3+CaCl_2=2NaCl+CaCO_3\)↓
0,25_______11/37__0,5______0,25
dd sau phản ứng gồm \(NaCl\)  \(0,5mol\) và \(CaCl_2\) dư = \(\dfrac{7}{148}\)mol

m dd sau phản ứng = trước phản ứng - m↓= 265 + 500 - \(0,25\times100\) = 740 (g)
=> C% dd \(NaCl\) = \(0,5\times\dfrac{58,5}{740}\times100\%\)  = 3,95%
C% dd \(CaCl_2\)= \(\dfrac{7}{148}\times\dfrac{111}{740}\times100\%\) = 0,71%

Bài 1:

a, Hiện tượng: Có khí mùi hắc thoát ra.

b, Ta có: \(m_{H_2SO_4}=100.24,5\%=24,5\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

PT: \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)

Theo PT: \(n_{Na_2SO_3}=n_{H_2SO_4}=0,25\left(mol\right)\)

\(\Rightarrow C\%_{Na_2SO_3}=\dfrac{0,25.126}{200}.100\%=15,75\%\)

c, Theo PT: \(n_{SO_2}=n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)

⇒ m _{dd sau pư} = 200 + 100 - 0,25.64 = 284 (g)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,25.142}{284}.100\%=12,5\%\)

Bài 2:

a, Hiện tượng: Xuất hiện kết tủa trắng.

PT: \(BaCl_2+MgSO_4\rightarrow MgCl_2+BaSO_{4\downarrow}\)

b, Ta có: \(m_{BaCl_2}=200.20,8\%=41,6\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{41,6}{208}=0,2\left(mol\right)\)

Theo PT: \(n_{MgCl_2}=n_{BaSO_4}=n_{MgSO_4}=n_{BaCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{ddMgSO_4}=\dfrac{0,2.120}{12\%}=200\left(g\right)\)

c, Ta có: m _{dd sau pư} = 200 + 200 - 0,2.233 = 353,4 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,2.95}{353,4}.100\%\approx5,38\%\)

Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

PTHH: MgO + 2HCl ---> MgCl₂ + H₂.

Theo PT: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)

=> \(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)

Ta có: \(m_{dd_{MgCl_2}}=4+100=104\left(g\right)\)

=> \(C_{\%_{MgCl_2}}=\dfrac{9,5}{104}.100\%=9,13\%\)

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

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

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

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

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

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%