`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2`

`0,1`    `0,2`            `0,1`        `0,1`             `(mol)`

`n_[HCl]=0,2.1=0,2(mol)`

 `=>m_[Zn]=0,1.65=6,5(g)`

`b)m_[dd HCl]=1,1.200=220(g)`

`=>C%_[ZnCl_2]=[0,1.136]/[6,5+220-0,1.2].100~~6%`

\(a,n_{HCl}=0,2.1=0,2\left(mol\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

            0,1<--0,2------>0,1------->0,1

\(\rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)

\(b,m_{ddHCl}=200.1,1=220\left(g\right)\)

\(\rightarrow m_{dd}=220+6,5-0,1.2=226,3\left(g\right)\\ \rightarrow C\%_{ZnCl_2}=\dfrac{0,1.136}{226,3}.100\%=6\%\)

\(a,PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\\ b,n_{Na_2CO_3}=\dfrac{15,9}{106}=0,15\left(mol\right)\\ \Rightarrow n_{HCl}=0,3\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,3\cdot36,5=10,95\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{10,95}{200}\cdot100\%=5,475\%\\ c,n_{CO_2}=0,15\left(mol\right)\\ \Rightarrow V_{CO_2\left(đkc\right)}=0,15\cdot24,79=3,7185\left(l\right)\\ d,m_{CO_2}=0,15\cdot44=6,6\left(g\right)\\ n_{NaCl}=0,3\left(mol\right);n_{H_2O}=0,15\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CT_{NaCl}}=0,3\cdot58,5=17,55\left(g\right)\\m_{H_2O}=0,15\cdot18=2,7\left(g\right)\end{matrix}\right.\\ m_{dd_{NaCl}}=15,9+200-2,7-6,6=206,6\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{17,55}{206,6}\cdot100\%\approx8,49\%\)

Bài 4 : 

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

300ml = 0,3l

\(n_{HCl}=1.0,3=0,3\left(mol\right)\)

1) Pt : \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

              1            2              1            1

             0,1         0,3           0,1

2) Lập tỉ số so sánh : \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\)

                      ⇒ MgO phản ứng hết , HCl dư

                      ⇒ Tính toán dựa vào số mol của MgO

\(n_{MgCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{MgCl2}=0,1.95=9,5\left(g\right)\)

\(n_{HCl\left(dư\right)}=0,3-\left(0,1.2\right)=0,1\left(mol\right)\)

⇒ \(m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

\(m_{ddHCl}=1,14.300=342\left(g\right)\)

\(m_{ddspu}=4+342=346\left(g\right)\)

\(C_{MgCl2}=\dfrac{9,5.100}{346}=2,75\)⁰/₀

\(C_{HCl\left(dư\right)}=\dfrac{3,65.100}{346}=1,05\)⁰/₀

 Chúc bạn học tốt

\(a)Mg+2HCl\rightarrow MgCl_2+H_2\)
\(b)n_{Mg}=\dfrac{3}{24}=0,125mol\\ n_{HCl}=0,1.1=0,1mol\\ \Rightarrow\dfrac{0,125}{1}>\dfrac{0,1}{2}\Rightarrow Mg.dư\\ n_{H_2}=n_{MgCl_2}=\dfrac{0,1}{2}=0,05mol\\ V_{H_2}=0,05.24,79=1,2395l\\ c)C_{M_{MgCl_2}}=\dfrac{0,05}{0,1}=0,5M\)

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow V_{H_2}=0,45.22,4=10,08\left(l\right)\)

c, Ta có: m _{dd sau pư} = 8,1 + 200 - 0,45.2 = 207,2 (g)

Theo PT: \(n_{AlCl_3}=n_{Al}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,3.133,5}{207,2}.100\%\approx19,33\%\)

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

a)\(Fe+2HCl\rightarrow FeCl_2+H_2\)

   0,15    0,3          0,15      0,15

   \(V_{H_2}=0,15\cdot22,4=3,36l\)

b)\(m_{H_2}=0,15\cdot2=0,3g\)

   \(BTKL:m_{ddFeCl_2}=8,4+100-0,3=108,1g\)

   \(m_{ctFeCl_2}=0,15\cdot127=19,05g\)

   \(C\%=\dfrac{m_{ctFeCl_2}}{m_{ddFeCl_2}}\cdot100\%=\dfrac{19,05}{108,1}\cdot100\%=17,62\%\)

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

PTHH: HCl + KOH → KCl + H₂O

Mol:     0,4      0,4        0,4

b) \(V_{ddKOH}=\dfrac{0,4}{1,5}=\dfrac{4}{15}\left(l\right)\approx0,267\left(l\right)\)

c) \(C_{M_{ddKCl}}=\dfrac{0,4}{0,2+\dfrac{4}{15}}=\dfrac{6}{7}M\approx0,857M\)

a) nHCl=0,2.2=0,4(mol

PTHH: HCl + KOH → KCl + H₂O

Mol:     0,4      0,4        0,4

b) VddKOH=0,4\1,5=4\15(l)≈0,267(l)

c) CMddKCl=0,4\(0,2+4\15)=6\7M≈0,857M