a) \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

\(n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

             0,25      0,5       0,5        0,5

Xét tỉ lệ : \(\dfrac{0,3}{1}>\dfrac{0,5}{2}\) => Zn dư , HCl đủ 

b) \(m_{Zn\left(dư\right)}=\left(0,3-0,25\right).65=3,25\left(g\right)\)

c) \(m_{ZnCl_2}=0,25.136=34\left(g\right)\)

\(V_{H_2}=0,25.22,4=5,6\left(l\right)\)

\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ a,Zn+2HCl\rightarrow ZnCl_2+H_2\\b, Vì:\dfrac{0,5}{2}< \dfrac{0,3}{1}\Rightarrow Zndư\\ n_{Zn\left(dư\right)}=0,3-\dfrac{0,5}{2}=0,05\left(mol\right)\\ \Rightarrow m_{Zn\left(dư\right)}=0,05.65=3,25\left(g\right)\\ c,n_{ZnCl_2}=n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,25.136=34\left(g\right)\\ V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

\(a,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ \text{Vì }\dfrac{n_{Zn}}{1}< \dfrac{n_{HCl}}{2}\text{ nên sau p/ứ }HCl\text{ dư}\\ \Rightarrow n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,2\cdot22,4=4,48\left(l\right)\\ b,\text{Chất còn dư là }HCl\\ n_{HCl\left(dư\right)}=n_{HCl\text{ đề}}-n_{HCl\text{ phản ứng}}=0,5-0,4=0,1\left(mol\right)\\ \Rightarrow m_{HCl\text{ dư}}=0,1\cdot36,5=3,65\left(g\right)\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

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

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 6,5 + 100 - 0,1.2 = 106,3 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)

Em cảm ơn idol

1.

a, \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:     0,15     0,3

b, Ta có: \(\dfrac{0,15}{1}< \dfrac{0,5}{2}\) ⇒ Mg pứ hết, HCl dư

\(m_{HCldư}=\left(0,5-0,3\right).36,5=7,3\left(g\right)\)

c, \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

2.

a, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

PTHH: 4P + 5O₂ ---t^o→ 2P₂O₅

Mol:     0,2    0,25              0,1

b, \(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

c, \(m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

d, \(V_{kk}=5,6.5=28\left(l\right)\)

n_Zn=\(\dfrac{13}{65}=0,2\left(mol\right)\)

n_HCl= 2 (mol) (theo đề bài)

a) PTHH:

Zn + 2HCl ----> ZnCl₂ + H₂^\(\uparrow\)

^{0,2.........2..................................................(mol)}

b)Tỉ lệ:\(\dfrac{0,2}{1}< \dfrac{2}{2}\) \(\rightarrow\) HCl dư (Tính theo n_Zn= 0,2 mol)

Theo PTHH: n_{HCl (pứ)} =\(\dfrac{0,2.2}{1}\) =0,4 (mol)

\(\rightarrow\) n_{HCl (dư)} = 2 - 0,4 = 1,6 (mol)

\(\rightarrow\) m_{HCl (dư)} = 1,6 . 36,5= 58,4(gam)

c) Theo PTHH: n_{\(H_2\)} = \(\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(\rightarrow\) V_{\(H_2\)}= 0,2 . 22,4 = 4,48 (l)

C.ơn bạn rất nhiều

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

Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\n_{HCl}=0,25\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,25}{2}\) \(\Rightarrow\) HCl còn dư, Kẽm p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,1\left(mol\right)\\n_{HCl\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=22,4\cdot0,1=2,24\left(l\right)\\m_{HCl\left(dư\right)}=0,05\cdot36,5=1,825\left(g\right)\end{matrix}\right.\)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\\ LTL:\dfrac{0,2}{1}< \dfrac{0,5}{2}\Rightarrow HCldư\\ n_{HCl\left(pứ\right)}=2n_{Zn}=0,4\left(mol\right)\\\Rightarrow m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65\left(g\right)\\ b.n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\\ c.n_{H_2}=n_{Zn}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4,=4,48\left(l\right)\\ d.3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O \\ n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12\left(mol\right)\\ LTL:\dfrac{0,2}{3}< \dfrac{0,12}{1}\Rightarrow Fe_2O_3dưsauphảnứng\\ \Rightarrow n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{15}\left(mol\right)\\ \Rightarrow m_{Fe}=\dfrac{2}{15}.56=7,467\left(g\right)\)

a) n\(Zn\)=\(\dfrac{m}{M}\)=\(\dfrac{13}{65}\)=0,2(mol)

n\(HCl\)=\(\dfrac{m}{M}\)=\(\dfrac{18,25}{36,5}=\)0,5(mol)

PTHH : Zn + 2HCl->ZnCl\(2\) + H\(2\)

            0,2     0,5

Lập tỉ lệ mol : \(^{\dfrac{0,2}{1}}\)<\(\dfrac{0,5}{2}\)

n\(Zn\) hết , n\(HCl\) dư

-->Tính theo số mol hết

               Zn + 2HCl->ZnCl\(2\) + H\(2\)

              0,2 ->  0,4      0,2        0,2

n\(HCl\) dư= n\(HCl\)(đề) - n\(HCl\)(pt)= 0,5 - 0,4 = 0,1(mol)

m​\(HCl\) dư= 0,1.36,5 = 3,65(g)

b) m\(ZnCl2\) = n.M= 0,2.136= 27,2 (g)

c)V\(H2\)=n.22,4=0,2.22,4=4,48(l)

d) n\(Fe\)\(2\)O\(3\)=\(\dfrac{m}{M}\)=\(\dfrac{19,2}{160}\)=0,12 (mol)

 3H2 +Fe2O3 → 2Fe + 3H2O

  0,2       0,12

Lập tỉ lệ mol: \(\dfrac{0,2}{3}\)<\(\dfrac{0,12}{1}\)

    nH2 hết .Tính theo số mol hết

\(HCl\)

​ 3H2 +Fe2O3 → 2Fe + 3H2O

0,2->                  0,2

m\(Fe\)=n.M= 0,2.56= 11,2(g)

a) PTHH: \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{BaCO_3}=\dfrac{68,95}{197}=0,35\left(mol\right)\\n_{HCl}=0,25\cdot3,2=0,8\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,35}{1}< \dfrac{0,8}{2}\) \(\Rightarrow\) Axit còn dư 

\(\Rightarrow n_{HCl\left(dư\right)}=0,8-0,35\cdot2=0,1\left(mol\right)\) \(\Rightarrow m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\)

b+c) Theo PTHH: \(\left\{{}\begin{matrix}n_{CO_2}=n_{BaCl_2}=0,35\left(mol\right)\\n_{HCl\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=0,35\cdot22,4=7,84\left(l\right)\\C_{M_{BaCl_2}}=\dfrac{0,35}{0,25}=1,4\left(M\right)\\C_{M_{HCl\left(dư\right)}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\end{matrix}\right.\) 

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

300ml = 0,3l

\(n_{HCl}=2.0,3=0,6\left(mol\right)\)

Pt : \(2Al+6HCl\rightarrow2AlCl_3+3H_2|\)

        2           6               2          3

       0,3       0,6

Lập tỉ số so sánh : \(\dfrac{0,3}{2}>\dfrac{0,6}{6}\)

             ⇒ Al dư , HCl phản ứng hết 

             ⇒ Tính toán dựa vào số mol của HCl

\(n_{Al\left(dư\right)}=0,3-\left(\dfrac{0,6.2}{6}\right)=0,1\left(mol\right)\)

⇒ \(m_{Al\left(dư\right)}=0,1.27=2,7\left(g\right)\)

 Chúc bạn học tốt

\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right);n_{HCl}=2.0,3=0,6\left(mol\right)\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:     0,2      0,6

Ta có: \(\dfrac{0,3}{2}>\dfrac{0,6}{6}\) ⇒ Al dư, HCl hết

\(m_{Aldư}=\left(0,3-0,2\right).27=2,7\left(g\right)\)