a) Zn + 2HCl --> ZnCl₂ + H₂

Hiện tượng: Kẽm tan dần, sủi bọt khí

b)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\); \(n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

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

c)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,1------------>0,1--->0,1

=> V_H2 = 0,1.22,4 = 2,24 (l)

m_ZnCl2 = 0,1.136 = 13,6 (g)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,1}=6\left(M\right)\)

Zn+2HCl->ZnCl₂+H₂

m_HCl=27,2+0,4-13=27,6-13=14,6(g)

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Áp.dụng.ĐLBTKL,ta.có:\\ m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\\ m_{H_2}=m_{Zn}+m_{HCl}-m_{ZnCl_2}=6,5+7,3-13,6=0,2\left(g\right)\)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)

\(n_{ZnCl_2}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(b)V_{ddHCl}=\dfrac{0,4}{2}=0,2l\\ m_{ZnCl_2}=0,2.136=27,2g\)

Câu 1:

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}=0,25\left(mol\right)\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)

c, \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\Rightarrow m_{Zn}=0,5.36,5=18,25\left(g\right)\)

Câu 2:

a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

b, \(n_P=\dfrac{1,24}{31}=0,04\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,05\left(mol\right)\Rightarrow V_{O_2}=0,05.22,4=1,12\left(l\right)\)

c, \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,02\left(mol\right)\Rightarrow m_{P_2O_5}=0,02.142=2,84\left(g\right)\)

1.

`Zn+2HCl->ZnCl_2+H_2`

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

`n_Zn=(16,25)/65=0,25 mol`

`=>V_(H_2)=0,25.22,4=5,6l`

`=>m_(HCl)=0,5.36,5=18,25g`

2.

`4P+5O_2->2P_2O_5`(to)

0,04---0,05----0,02

`nP=(1,24)\31=0,04 mol`

`V_(O_2)=0,05.22,4=1,12l`

`m_(P_2O_5)=0,02.142=2,84g`

\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)

a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)

    0,4           0,6                0,2             0,6

b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

=>\(n_{Al}=0.4\left(mol\right)\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

c: \(4Al+3O_2\rightarrow2Al_2O_3\)

0,4                      0,2

\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)

\(n_{HCl}=\dfrac{25}{36,5}=\dfrac{50}{73}mol\)

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

\(\Rightarrow n_{Al}=\dfrac{\dfrac{50}{73}.2}{6}=\dfrac{50}{219}mol\\ m_{Al}=\dfrac{50}{219}.27=\dfrac{450}{73}g\)

\(n_{H_2}=\dfrac{\dfrac{50}{73}.3}{6}=\dfrac{25}{73}mol\\ V_{H_2}=\dfrac{25}{73}.22,4=\dfrac{560}{73}l\)

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

b: \(n_{HCl}=\dfrac{25}{36.5}=\dfrac{50}{73}\left(mol\right)\)

\(\Leftrightarrow n_{AlCl_3}=\dfrac{150}{73}\left(mol\right)=n_{Al}\)

\(m_{Al}=\dfrac{150}{73}\cdot27=\dfrac{4050}{73}\left(g\right)\)

\(n_{Zn}=\dfrac{32,5}{65}=0,5(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{H_2}=n_{Zn}=0,5(mol)\\ \Rightarrow V_{H_2(phản ứng)}=0,5.22,4=11,2(l)\\ \Rightarrow V_{H_2(thực tế)}=11,2.80\%=8,96(l)\)