1.\(m_{BaCl_2}\) = 200.25% = 50g

2.\(C\%_{Cu\left(NO_3\right)_2}\) = \(\dfrac{40}{200}.100\) = \(20\%\)

3.Ở 25\(^o\)C 100g nước hòa tan được 36 g NaCl để tạo dung dịch bão hòa

\(m_{dd}=m_{ct}+m_{dm}=36+100=136g\)

Nồng độ phần trăm của dung dịch NaCl:C% = \(\dfrac{36}{136}.100=26,47\%\)

\(1,m_{BaCl_2}=\dfrac{200.25}{100}=50\left(g\right)\\ 2,C\%_{Cu\left(NO_3\right)_2}=\dfrac{4}{200}.100\%=20\%\\ 3,C\%=\dfrac{36}{36+100}.100\%=26,47\%\)

Câu trả lời đúng: B.

mdd = mct + mnước ⇒ mnước = mdd - mct = 200g – 10g = 190g

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH:

Fe + H₂SO₄ ---> FeSO₄ + H₂

0,1         0,1          0,1        0,1

\(\rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\\ \rightarrow m_{Fe_3O_4}=7,92-5,6=2,32\left(g\right)\\ \rightarrow n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)

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

0,01         0,04              0,01             0,01

\(n_{H_2SO_4\left(pư\right)}=0,1+0,04=0,14\left(mol\right)\)

\(m_{H_2SO_4\left(bđ\right)}=200.20\%=40\left(g\right)\)

\(\rightarrow n_{H_2SO_4\left(bđ\right)}=\dfrac{40}{98}=0,41\left(mol\right)\)

So sánh: 0,41 > 0,14 => H₂SO₄ có dư

=> \(n_{H_2SO_4\left(dư\right)}=0,41-0,14=0,27\left(mol\right)\)

\(n_{FeSO_4}=0,1+0,01=0,11\left(mol\right)\)

PTHH:

H₂SO₄ + BaCl₂ ---> BaSO₄ + 2HCl

0,27                          0,27

FeSO₄ + BaCl₂ ---> FeCl₂ + BaSO₄

0,11                          0,11       0,11

Fe₂(SO₄)₃ + 3BaCl₂ ---> 2FeCl₃ + 3BaSO₄

0,1                                     0,2            0,3

\(\rightarrow\left\{{}\begin{matrix}m_{BaSO_4}=\left(0,3+0,27+0,11\right).158,44\left(g\right)\\m_{FeCl_2}=0,11.127=13,97\left(g\right)\\m_{FeCl_3}=0,1.162,5=16,25\left(g\right)\end{matrix}\right.\)

=> m_muối = 158,44 + 13,97 + 16,25 = 188,66 (g)

1)

$3AgNO_3 + AlCl_3 \to Al(NO_3)_3 + 3AgCl$
$K_2SO_4 + BaCl_2 \to BaSO_4 + 2KCl$
$6KOH + Al_2(SO_4)_3 \to 2Al(OH)_3 + 3KCl$
$Na_2CO_3 + 2HCl \to 2NaCl + CO_2 + H_2O$

2_

$CuCl_2 + 2NaOH \to Cu(OH)_2 + 2NaCl$

Theo PTHH :

$n_{CuCl_2} = n_{Cu(OH)_2} = \dfrac{1}{2}n_{NaOH} = \dfrac{1}{2}.\dfrac{200.20\%}{40} = 0,25(mol)$
$m_{CuCl_2} = 0,25.135 = 33,75(gam)$
$m_{Cu(OH)_2} = 0,25.98 = 24,5(gam)$

Bài 10:

PTHH: \(Na_2CO_3+BaCl_2\rightarrow2NaCl+BaCO_3\downarrow\)

a) Ta có: \(n_{Na_2CO_3}=\dfrac{200\cdot10,6\%}{106}=0,2\left(mol\right)=n_{BaCO_3}\)

\(\Rightarrow m_{BaCO_3}=0,2\cdot197=39,4\left(g\right)\)

b) Theo PTHH: \(n_{BaCl_2}=n_{BaCO_3}=0,2mol\)

\(\Rightarrow C\%_{BaCl_2}=\dfrac{0,2\cdot208}{120}\cdot100\%\approx34,67\%\)

c) Theo PTHH: \(n_{NaCl}=2n_{BaCl_2}=0,4mol\) \(\Rightarrow m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)

Mặt khác: \(m_{dd}=m_{ddNa_2CO_3}+m_{ddBaCl_2}-m_{BaCO_3}=280,6\left(g\right)\)

\(\Rightarrow C\%_{NaCl}=\dfrac{23,4}{280,6}\cdot100\%\approx8,34\%\)

Bài 11 và bài 12 tương tự bài 10

\(n_{Na_2CO_3}=\dfrac{200.13,25%}{100\%.106}=0,25(mol)\\ n_{BaCl_2}=\dfrac{20,8\%.300}{100\%.208}=0,3(mol)\\ PTHH:Na_2CO_3+BaCl_2\to BaSO_4\downarrow +2NaCl\)

Vì \(\dfrac{n_{Na_2CO_3}}{1}<\dfrac{n_{BaCl_2}}{1}\) nên \(BaCl_2\) dư

Do đó dd sau p/ứ gồm \(BaCl_2\) dư và \(NaCl\)

\(n_{BaSO_4}=0,25(mol);n_{NaCl}=0,5(mol);n_{BaCl_2(dư)}=0,3-0,25=0,05(mol)\\ \Rightarrow m_{NaCl}=0,5.58,5=29,25(g);m_{BaCl_2(dư)}=0,05.208=10,4(g)\\ \Rightarrow C\%_{NaCl}=\dfrac{29,25}{200+300-0,25.233}.100\%=6,62\%\\ C\%_{BaCl_2(dư)}=\dfrac{10,4}{200+300-0,25.233}.100\%=2,35\%\)