a. Ta có: \(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

Ta lại có: \(C_{\%_{HCl}}=\dfrac{m_{ct_{HCl}}}{100}.100\%=7,3\%\)

=> m_HCl = 7,3(g)

=> \(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

PTHH:

Fe₃O₄ + 8HCl ---> FeCl₂ + 2FeCl₃ + 4H₂O

    1   --->   8

    0,1 --->  0,2

=> \(\dfrac{0,1}{1}>\dfrac{0,2}{8}\)

Vậy Fe₃O₄ dư

=> m_dư = 23,2 - 7,3 = 15,9 (g)

b. Theo PT: \(n_{FeCl_2}=\dfrac{1}{8}.n_{HCl}=\dfrac{1}{8}.0,2=0,025\left(mol\right)\)

=> \(m_{FeCl_2}=0,025.127=3,175\left(g\right)\)

Theo PT: \(n_{FeCl_3}=\dfrac{1}{4}.n_{HCl}=\dfrac{1}{4}.0,2=0,05\left(mol\right)\)

=> \(m_{FeCl_3}=0,05.162,5=8,125\left(g\right)\)

=> \(m_{muối}=8,125+3,175=11,3\left(g\right)\)

c. Ta có: m_{dung dịch sau PỨ} = \(23,2+100=123,2\left(g\right)\)

Theo PT: \(n_{H_2O}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

=> \(m_{H_2O}=0,1.18=1,8\left(g\right)\)

m_{các chất sau PỨ} = 1,8 + 11,3 = 13,1(g)

=> \(C_{\%_{sauPỨ}}=\dfrac{13,1}{123,2}.100\%=10,63\%\)

\(n_{NaOH}=\dfrac{200.5\%}{100\%.40}=0,25(mol)\\ n_{HCl}=\dfrac{36,5.20\%}{100\%.36,5}=0,2(mol)\\ a,NaOH+HCl\to NaCl+H_2O\)

Vì \(\dfrac{n_{NaOH}}{1}>\dfrac{n_{HCl}}{1}\) nên \(NaOH\) dư

\(b,n_{NaOH(dư)}=0,25-0,2=0,05(mol);n_{NaCl}=0,2(mol)\\ \Rightarrow m_{\text{dd sau p/ứ}}=0,2.58,5+0,05.40=13,7(g)\\ c,m_{NaCl}=0,2.58,5=11,7(g)\\ d,n_{H_2}=0,2(mol)\\ \Rightarrow \begin{cases} C\%_{NaOH}=\dfrac{0,05.40}{36,5+200-0,2.2}.100\%=0,85\%\\ C\%_{NaCl}=\dfrac{11,7}{36,5+200-0,2.2}.100\%=4,96\% \end{cases}\)

a) \(n_{CO_2}=\dfrac{0,4958}{24,79}=0,02\left(mol\right);n_{HCl}=0,6.1=0,6\left(mol\right)\)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

            0,02<------0,04<----0,02<-----0,02

\(\Rightarrow n_{HCl\left(p\text{ư}\right)}< n_{HCl\left(b\text{đ}\right)}\left(0,04< 0,6\right)\Rightarrow HCl\) dư, \(CaCO_3\) tan hết

\(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,02.100=2\left(g\right)\\m_{CaSO_4}=5-2=3\left(g\right)\end{matrix}\right.\)

b) dd sau phản ứng có: \(\left\{{}\begin{matrix}n_{HCl\left(d\text{ư}\right)}=0,6-0,04=0,56\left(mol\right)\\n_{CaCl_2}=0,02\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M\left(HCl\left(d\text{ư}\right)\right)}=\dfrac{0,56}{0,6}=\dfrac{14}{15}M\\C_{M\left(CaCl_2\right)}=\dfrac{0,02}{0,6}=\dfrac{1}{30}M\end{matrix}\right.\)

Gọi n_Na2CO3 = x (mol)

Na₂CO₃ + 2HCl \(\rightarrow\) 2NaCl + H₂O + CO₂

  x         \(\rightarrow\)    2x   \(\rightarrow\)    2 x                                    (mol)

C%_(NaCl) = \(\frac{2.58,5x}{200+120}\) . 100% = 20%

=> x =0,547 (mol)

m_Na2CO3 = 0,547 . 106 = 57,982 (g)

m_HCl = 2 . 0,547 . 36,5 =39,931 (g)

C%_(Na2CO3) =\(\frac{57,892}{200}\) . 100% = 28,946%

C%(HCl) = \(\frac{39,931}{120}\) . 100% = 33,28%

\(m_{ct}=\dfrac{5.200}{100}=10\left(g\right)\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)

           1             1             1         1

        0,25          0,25        0,25

a) \(n_{HCl}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

\(m_{HCl}=0,25.36,5=9,125\left(g\right)\)

\(m_{ddHCl}=\dfrac{9,125.100}{3,65}=250\left(g\right)\)

b) \(n_{NaCl}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

⇒ \(m_{NaCl}=0,25.58,5=14,625\left(g\right)\)

\(m_{ddspu}=200+250=450\left(g\right)\)

\(C_{NaCl}=\dfrac{14,625.100}{450}=3,25\)⁰/₀

 Chúc bạn học tốt

a) \(n_{NaOH}=\dfrac{200.5\%}{40}=0,25\left(mol\right)\)

PTHH: NaOH + HCl → NaCl + H₂O

Mol:       0,25      0,25    0,25

\(m_{ddHCl}=\dfrac{0,25.36,5.100}{3,65}=250\left(g\right)\)

b) m_{dd sau pứ} = 200 + 250 = 450 (g)

\(C\%_{ddNaCl}=\dfrac{0,25.58,5.100\%}{450}=3,25\%\)

a) $NaOH + HCl \to NaCl + H_2O$

$n_{HCl} = n_{NaOH} = \dfrac{200.5\%}{40} = 0,25(mol)$

$\Rightarrow n_{HCl} = \dfrac{0,25.36,5}{3,65\%} = 250(gam)$

b)

$m_{dd\ sau\ pư} = 200 + 250 = 450(gam)$
$C\%_{NaCl} = \dfrac{0,25.58,5}{450}.100\% = 3,25\%$

Bài 1 :

PTHH : Zn + H₂SO₄ ------> ZnSO₄ + H₂

\(n_{Zn}=\frac{m}{M}=\frac{26}{65}=0,4\left(mol\right)\)

Theo PTHH : n_H2 = n_Zn = 0,4 mol

=> Khối lượng H₂ được tạo ra ở đktc là :

\(V=n\times22,4\)

\(\Rightarrow V_{H_2}=0,4\times22,4\)

\(\Rightarrow V_{H_2}=8,96\left(l\right)\)

Theo PTHH : n_ZnSO4 = n_Zn = 0,4 mol

=> Khối lượng muối được tạo thành là :

\(m=n\times M\)

\(\Rightarrow m_{ZnSO_4}=0,4\times161\)

\(\Rightarrow m_{ZnSO_4}=64,4\left(g\right)\)

b) Theo PT : \(n_{H_2SO_4}=n_{Zn}=0,4\left(mol\right)\)

=> Khối lượng \(H_2SO_4\)cần dùng cho phản ứng là :

\(m=n\times M\)

\(\Rightarrow m_{H_2SO_4}=0,4\times98\)

\(\Rightarrow m_{H_2SO_4}=39,2\left(g\right)\)

c) Nồng độ phần trăm thu được sau phản ứng là :

\(C\%=\frac{m_{ct}}{m_{dd}}\times100\%\)

\(\Rightarrow C\%=\frac{39,2}{64,4}\times100\%\approx60,9\%\)

Vậy :.........................

\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ n_{ZnCl_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ m_{ZnCl_2} = 0,1.136 = 13,6(gam)\\ b) n_{H_2} = n_{Zn} = 0,1(mol) \Rightarrow V_{H_2} = 0,1.22,4 =2 ,24(lít)\\ c) n_{HCl} =2 n_{H_2} = 0,2(mol)\\ \Rightarrow m_{HCl} = 0,2.36,5 = 7,3(gam)\ ; V_{dd\ HCl} = \dfrac{0,2}{0,5} = 0,4(lít)\)

a)

$K_2SO_4 + BaCl_2 \to BaSO_4 + 2KCl$

b)

$n_{K_2SO_4} = 0,2.2 = 0,4(mol)$
$n_{BaCl_2} = 0,3.1 = 0,3(mol)$

Ta thấy : 

$n_{K_2SO_4} : 1 > n_{BaCl_2} : 1$ nên $K_2SO_4$ dư

$n_{BaSO_4} = n_{BaCl_2} = 0,3(mol)$
$m_{BaSO_4} = 0,3.233 = 69,9(gam)$

c) $n_{K_2SO_4} = 0,4 - 0,3 = 0,1(mol)$

$V_{dd\ sau\ pư} = 0,2 + 0,3 = 0,5(lít)$

$C_{M_{K_2SO_4} } = \dfrac{0,1}{0,5} = 0,2M$
$C_{M_{KCl}} = \dfrac{0,6}{0,5} = 1,2M$

\(a.n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{17,8\%.200}{36,5}=\dfrac{356}{365}\left(mol\right)\\ Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{\dfrac{356}{365}}{2}\\ \Rightarrow Znhết,HCldư\\ n_{HCl\left(dùng\right)}=0,1.2=0,2\left(mol\right)\\ m_{HCl\left(dùng\right)}=0,2.36,5=7,3\left(g\right)\\ b.n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c.n_{HCl\left(Dư\right)}=\dfrac{356}{365}-0,2=\dfrac{283}{365}\left(mol\right)\\ C\%_{ddZnCl_2}=\dfrac{0,1.136}{6,5+200}.100\approx6,586\%\)

\(C\%_{ddHCl\left(dư\right)}=\dfrac{\dfrac{283}{365}.36,5}{6,5+200}.100\approx13,705\%\)