200ml = 0,2l

300ml = 0,3l

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

\(n_{HCl}=0,8.0,3=0,24\left(mol\right)\)

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

               1            1            1            1

              0,2          0,24      0,2

b) Lập tỉ số so sánh : \(\dfrac{0,2}{1}< \dfrac{0,24}{1}\)

                ⇒ NaOH phản ứng hết , HCl dư

                ⇒ Tính toán dựa vào số mol của NaOH

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

\(n_{HCl\left(dư\right)}=0,24-0,2=0,04\left(mol\right)\)

\(V_{ddspu}=0,2+0,3=0,5\left(l\right)\)

\(C_{M_{NaCl}}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

\(C_{M_{HCl\left(dư\right)}}=\dfrac{0,04}{0,5}=0,08\left(M\right)\)

c) Cho quỳ tím vào dung dịch A , quỳ tím sẽ chuyển sang màu đỏ 

 Chúc bạn học tốt

\(n_{NaOH}=0,2.1=0,2\left(mol\right);n_{HCl}=0,3.0,8=0,24\left(mol\right)\\ a.NaOH+HCl\rightarrow NaCl+H_2O\\b. Vì:\dfrac{0,2}{1}< \dfrac{0,24}{1}\\ \Rightarrow HCldư\\ n_{NaCl}=n_{NaOH}=n_{HCl\left(p.ứ\right)}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,24-0,2=0,04\left(mol\right)\\ V_{ddsau}=0,2+0,3=0,5\left(l\right)\\ C_{MddHCl\left(dư\right)}=\dfrac{0,04}{0,5}=0,08\left(M\right)\\ C_{MddNaCl}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)

a) n HCl = 0,2.2,5 = 0,5 mol

b) m HCl =200.7,3% = 14,6 gam

n HCl = 14,6/36,5 = 0,4 mol

c) m NaOH = 300.40% = 120 gam

n NaOH = 120/40 = 3(mol)

d) n NaOH = 0,5.0,5 = 0,25 mol

MgCl2+2AgNO3->Mg(NO3)2+2AgCl

0,04-----0,08-----------0,04----------0,08

n MgCl2=0,1 mol

n AgNO3=0,08 mol

=>Mgcl2 dư

=>m AgCl=0,08.143,5=11,48g

=>CMMg(NO)2=\(\dfrac{0,04}{0,2}\)=0,2M

=>CMMgcl2 dư=\(\dfrac{0,06}{0,2}\)=0,3M

\(n_{MgCl_2}=0,1\cdot1=0,1mol\)

\(n_{AgNO_3}=0,1\cdot0,8=0,08mol\)

\(MgCl_2+2AgNO_3\rightarrow2AgCl\downarrow+Mg\left(NO_3\right)_2\)

0,1            0,08                 0              0

0,04          0,08                 0,08         0,04

0,06          0                      0,08         0,04

\(m_{\downarrow}=0,08\cdot143,5=11,48g\)

\(C_{M_{Mg\left(NO_3\right)_2}}=\dfrac{n_{Mg\left(NO_3\right)_2}}{V_X}=\dfrac{0,04}{0,2}=0,2M\)

\(n_{MgCl_2}=0,1.1=0,1\left(mol\right)\\ n_{AgNO_3}=0,1.0,8=0,08\left(mol\right)\)

PTHH: MgCl₂ + 2AgNO₃ ---. 2AgCl + Mg(NO₃)₂

LTL: \(0,1>\dfrac{0,08}{2}\rightarrow\) MgCl₂ dư

Theo pthh: \(\left\{{}\begin{matrix}n_{MgCl_2\left(pư\right)}=n_{Mg\left(NO_3\right)_2}=\dfrac{0,08}{2}=0,04\left(mol\right)\\n_{AgCl}=0,08\left(mol\right)\end{matrix}\right.\)

\(\rightarrow C_M_{MgCl_2\left(dư\right)}==\dfrac{0,1-0,04}{0,2}=0,3M\\ C_{MMg\left(NO_3\right)_2}=\dfrac{0,04}{0,2}=0,2M\)

\(m_{AgCl}=143,5.0,08=11,48\left(g\right)\)

n_HCl = 0.4 (mol) 

n_Ba(OH)2 = 0.15 (mol) 

Ba(OH)₂ + 2HCl => BaCl₂ + H₂O

0.15_______0.3_____0.15 

C_{M HCl(dư)} = 0.3/0.4 = 0.75M

C_{M BaCl2} = 0.15/(0.4 + 0.3) = 3/14 (M) 

Bài 1:

\(n_{KNO_3}=\dfrac{20}{101}=0,198\left(mol\right)\)

\(C_M=\dfrac{n}{V}=\dfrac{0,198}{0,85}=0,233M\)

Bài 2:

\(C_M=\dfrac{n}{V}=\dfrac{0,5}{0,75}=0,66M\)

Bài 3:

\(n_{KNO_3}=2.0,5=1\left(mol\right)\)

\(m_{KNO_3}=1.101=101\left(g\right)\)

Bài 4:

\(C\%=\dfrac{20}{600}.100=3,33\%\)

Bài 1:

\(n_{KNO_3}=\dfrac{20}{101}=0,198\left(mol\right)\)

\(C_{M_{ddKNO_3}}=\dfrac{0,198}{0,85}\approx0,23M\)

Bài 2:

\(C_{M_{ddKCl}}=\dfrac{0,5}{0,75}\approx0,667M\)

Bài 3:

\(n_{KNO_3}=0,5.2=1\left(mol\right)\Rightarrow m_{KNO_3}=1.101=101\left(g\right)\)

Bài 4:

\(C\%_{ddKCl}=\dfrac{20.100\%}{600}=3,333\%\)

Đáp án D

Bài 1:

\(a.n_{NaOH\left(tổng\right)}=0,05.1+0,2.0,2=0,09\left(mol\right)\\ V_{ddNaOH\left(tổng\right)}=50+200=250\left(ml\right)=0,25\left(l\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,09}{0,25}=0,36\left(M\right)\\ b.n_{HCl}=0,5.0,02=0,01\left(mol\right)\\ n_{H_2SO_4}=0,08.0,2=0,016\left(mol\right)\\ V_{ddsau}=20+80=100\left(ml\right)=0,1\left(l\right)\\ C_{MddH_2SO_4}=\dfrac{0,016}{0,1}=0,16\left(M\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\)

Bài 2:

\(a.m_{H_2SO_4}=29,4.10\%=2,94\left(g\right)\\ b.n_{H_2SO_4}=\dfrac{2,94}{98}=0,03\left(mol\right)\\ n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,03}{1}\Rightarrow H_2SO_4dư\\ n_{H_2SO_4\left(dư\right)}=0,03-0,01=0,02\left(mol\right)\\ m_{H_2SO_4\left(dư\right)}=0,02.98=1,96\left(g\right)\\ n_{H_2}=n_{Fe}=0,01\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\)

a, \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)=3360\left(ml\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{FeCl_2}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)

c, Ta có: \(m_{ddHCl}=1,25.200=250\left(g\right)\)

⇒ m _{dd sau pư} = 8,4 + 250 - 0,15.2 = 258,1 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,15.127}{258,1}.100\%\approx7,38\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,1.36,5}{258.1}.100\%\approx1,41\%\end{matrix}\right.\)