a/ \(n_{KOH}=0,2.1=0,2\left(mol\right);n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Mol:        0,2         0,1            0,1

Ta có: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) ⇒ KOH hết, H₂SO₄ dư

b/ \(m_{H_2SO_4dư}=\left(0,3-0,1\right).98=19,6\left(g\right)\)

c/ V_{dd sau pứ} = 0,2 + 0,3 = 0,5 (l)

d/ \(C_{M_{ddK_2SO_4}}=\dfrac{0,1}{0,5}=0,2M\)

 \(C_{M_{ddH_2SO_4dư}}=\dfrac{0,3-0,1}{0,5}=0,4M\)

PTHH: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)

            \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CuCl_2}=0,2\cdot2=0,4\left(mol\right)\\n_{NaOH}=0,2\cdot2=0,4\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) \(\Rightarrow\) CuCl₂ còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,4\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,2\left(mol\right)=n_{CuO}=n_{CuCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2\cdot80=16\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,2+0,2}=1\left(M\right)\\C_{M_{CuCl_2}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\\\end{matrix}\right.\)

a) \(n_{NaOH}=0,2.1=0,2\left(mol\right);n_{H_2SO_4}=0,15.2=0,3\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

Mol:         0,2                            0,1

Ta có: \(\dfrac{0,2}{2}< \dfrac{0,3}{1}\) ⇒ NaOH hết, H₂SO₄ dư

\(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

b) V_{dd sau pứ} = 0,2 + 0,15 = 0,35 (l)

\(C_{M_{ddNa_2SO_4}}=\dfrac{0,1}{0,35}=\dfrac{2}{7}\approx0,2857M\)

\(C_{M_{ddH_2SO_4dư}}=\dfrac{0,3-0,1}{0,35}=\dfrac{4}{7}\approx0,57M\)

\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)

\(n_{CuSO_4}=0.1\cdot2=0.2\left(mol\right)\)

\(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\)

\(0.1.............0.05...............0.05...........0.05\)

\(m_{Cu\left(OH\right)_2}=0.05\cdot98=4.9\left(g\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.05}{0.2+0.1}=0.167\left(M\right)\)

\(C_{M_{CuSO_4\left(dư\right)}}=\dfrac{0.2-0.05}{0.1}=1.5\left(M\right)\)

2NaOH + CuSO4 → Cu(OH)2 + Na2SO4

n NaOH = 0,2.5 = 1(mol)

n CuSO4 = 0,1.2 = 0,2(mol)

Ta có :

n NaOH / 2 = 0,5 > n CuSO4 / 1 = 0,2 => NaOH dư

n Cu(OH)2 = n CuSO4 = 0,2 mol

=> m A = 0,2.98 = 19,6 gam

n Na2SO4 = n CuSO4 = 0,2 mol

n NaOH pư = 2n CuSO4 = 0,4(mol)

V dd = 0,2 + 0,1 = 0,3(lít)

Suy ra:

CM Na2SO4 = 0,2/0,3 = 0,67M

CM NaOH = (1 - 0,4)/0,3 = 2M

\(n_{BaCl_2}=0.2\cdot0.5=0.1\left(mol\right)\)

\(BaCl_2+K_2SO_4\rightarrow BaSO_4+2KCl\)

\(0.1.............0.1.........................0.2\)

\(V_{dd_{K_2SO_4}}=\dfrac{0.1}{1}=0.1\left(l\right)\)

\(V_{dd}=0.2+0.1=0.3\left(l\right)\)

\(C_{M_{KCl}}=\dfrac{0.2}{0.3}=0.67\left(M\right)\)

Đổi 200ml = 0,2 lít

Ta có: \(n_{BaCl_2}=0,5.0,2=0,1\left(mol\right)\)

a. PTHH: \(BaCl_2+K_2SO_4--->BaSO_4\downarrow+2KCl\)

Theo PT: \(n_{K_2SO_4}=n_{BaCl_2}=0,1\left(mol\right)\)

\(\Rightarrow V_{dd_{K_2SO_4}}=\dfrac{0,1}{1}=0,1\left(lít\right)\)

b. Theo PT: \(n_{KCl}=2.n_{BaCl_2}=2.0,1=0,2\left(mol\right)\)

Ta có: \(V_{dd_{KCl}}=V_{dd_{BaCl_2}}=0,1\left(lít\right)\)

\(\Rightarrow C_{M_{KCl}}=\dfrac{0,2}{0,1}=2M\)

PTHH: \(CaCl_2+Na_2CO_3\rightarrow CaCO_3\downarrow+2NaCl\)

a+b) Ta có: \(n_{CaCl_2}=0,1\cdot2=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{CaCO_3}=0,2\left(mol\right)\\n_{NaCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=0,2\cdot100=20\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,4}{0,1+0,2}\approx1,33\left(M\right)\end{matrix}\right.\)

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

Theo PTHH: \(n_{HCl}=2n_{CaCO_3}=0,4\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{0,4\cdot36,5}{10\%}=146\left(g\right)\)

a.PTHH:CaCl2+Na2CO3--->CaCO3+2NaCl

Ta có:nCaCl2=0,2

=>nCaCO3=nCaCl2=0,2(mol)=>mCaCO3(kết tủa)=100.0,2=2(g)

b.Vdd=100+200=300(ml)=0,3(l)

CM Nacl=(2.0,2)/0,3=4/3(M)(Đề cho 2 chất td vừa đủ nên dd sau pứ chỉ có NaCl)

c.CaCO3+2HCl--->CaCl2+CO2+H2O

nHCl(cần dùng)=2.0.2=0,4(mol)=>mHCl=36,5.0,4=14,6(g)

=>mddHCl=14,6/10%=146(g)

\(n_{H_2SO_4}=1.0,2=0,2(mol)\\ H_2SO_4+BaCl_2\to BaSO_4\downarrow+2HCl\\ \Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,2(mol)\\ a,m_{BaSO_4}=0,2.233=46,6(g)\\ b,V_{dd_{BaCl_2}}=\dfrac{0,2}{1,5}\approx 0,13(l)\\ c,n_{HCl}=0,4(mol)\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,2+0,13}\approx 1,21M\)

\(d,\) Dd sau p/ứ là HCl nên làm quỳ tím hóa đỏ

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\\ n_{BaCl_2}=n_{BaSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\\ n_{HCl}=2.0,2=0,4\left(mol\right)\\ a,m_{\downarrow}=m_{BaSO_4}=0,2.233=46,6\left(g\right)\\ b,V_{\text{dd}BaCl_2}=\dfrac{0,2}{1,5}=\dfrac{2}{15}\left(l\right)\\ c,C_{M\text{dd}HCl}=\dfrac{0,4}{\dfrac{2}{15}+0,2}=1,2\left(M\right)\\ d,V\text{ì}.c\text{ó}.\text{dd}.HCl\Rightarrow Qu\text{ỳ}.ho\text{á}.\text{đ}\text{ỏ}\)

200ml = 0,2l

\(n_{H2SO4}=1.0,2=0,2\left(mol\right)\)

Pt : \(BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl|\)

           1           1                  1              2

          0,2        0,2               0,2           0,4

a) \(n_{BaSO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{BaSO4}=0,2.233=46,6\left(g\right)\)

b) \(n_{BaCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{ddBaCl2}=\dfrac{0,2}{1,5}=\dfrac{2}{15}\left(l\right)=133,33\left(ml\right)\)

c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

\(V_{ddspu}=0,2+\dfrac{2}{15}=\dfrac{1}{3}\left(l\right)\)

\(C_{M_{ddHCl}}=\dfrac{0,4}{\dfrac{1}{3}}=1,2\left(M\right)\)

d) Cho quỳ tím vào dung dịch sau phản ứng , quỳ tím hóa đỏ. Vì dung dịch thu được sau phản ứng có HCl là axit

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