A)n_Zn=6,5/65=0,1(m)

PTHH : Zn + 2HCl -> ZnCl₂ + H₂

tỉ lệ     :1       2           1           1

số mol:0,1     0,2      0,1         0,1

V\(_{H_2}\)=0,1.22,4=2,24(l)

B)m\(_{ZnCl_2}\)=0,1.136=13,6(g)

góc soi:

đã ghi công thức thì ghi luôn kí hiệu đi chứ:)))

a, \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

b, \(3H_2+Fe_2O_3\underrightarrow{t^o}2Fe+3H_2O\)

c, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

e, \(2K+2H_2O\rightarrow2KOH+H_2\)

f, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

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

b, \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)

Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,2\left(mol\right)\Rightarrow m_{P_2O_5}=0,2.142=28,4\left(g\right)\)

c, \(n_{O_2}=\dfrac{5}{4}n_P=0,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)

d, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}n_{O_2}=\dfrac{1}{3}\left(mol\right)\Rightarrow m_{KClO_3}=\dfrac{1}{3}.122,5=\dfrac{245}{6}\left(g\right)\)

mình cảm ơnn

a, \(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

b, \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

c, \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

d, \(2Na+\dfrac{1}{2}O_2\underrightarrow{t^o}Na_2O\)

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

f, \(2Mg+O_2\underrightarrow{t^o}2MgO\)

a, \(4Fe+3O_2\underrightarrow{t^o}2Fe_2O_3\)

b, \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

c, \(H_2O\underrightarrow{đp}H_2+\dfrac{1}{2}O_2\)

d, \(H_2+\dfrac{1}{2}O_2\underrightarrow{t^o}H_2O\)

e, \(C+O_2\underrightarrow{t^o}CO_2\)

f, \(C_3H_6+\dfrac{9}{2}O_2\underrightarrow{t^o}3CO_2+3H_2O\)

\(4Fe+3O_2-^{t^o}>2Fe_2O_3\\ 2KMnO_4-^{t^o}>K_2MnO_4+MnO_2+O_2\\ 2H_2O-^{điệnphân}>2H_2+O_2\\ 2H_2+O_2-^{t^o}>2H_2O\\ C+O_2-^{t^o}>CO_2\\ 2C_3H_6+9O_2-^{t^o}>6CO_2+6H_2O\)

a, Gọi KL cần tìm là A.

\(A+2HCl\rightarrow ACl_2+H_2\)

Ta có: \(n_A=\dfrac{4,8}{M_A}\left(mol\right)\), \(n_{ACl_2}=\dfrac{1}{M_A+71}\left(mol\right)\)

Theo PT: \(n_A=n_{ACl_2}\Rightarrow\dfrac{4,8}{M_A}=\dfrac{1}{M_A+71}\Rightarrow M_A=-89,68\)

→ vô lý

Bạn xem lại đề nhé.

a, \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b, Sửa đề: 8,4 (g) → 8,1 (g)

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

Xét tỉ lệ: \(\dfrac{0,3}{2}< \dfrac{0,5}{3}\), ta được H₂SO₄ dư.

Theo PT: 

\(n_{H_2SO_4\left(pư\right)}=\dfrac{3}{2}n_{Al}=0,45\left(mol\right)\Rightarrow n_{H_2SO_4\left(dư\right)}=0,5-0,45=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

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

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

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

PT: \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

Theo PT: \(n_{Ca\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}+n_{H_2SO_4}=0,4\left(mol\right)\)

\(\Rightarrow V_{ddCa\left(OH\right)_2}=\dfrac{0,4}{1}=0,4\left(l\right)\)

a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

b, Sửa đề: 17,9 (l) → 17,92 (l)

Ta có: \(n_{CO_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right)=n_C\)

\(n_{H_2O}=\dfrac{18}{18}=1\left(mol\right)\Rightarrow n_H=1.2=2\left(mol\right)\)

⇒ m_A = m_C + m_H = 0,8.12 + 2.1 = 11,6 (g)

Theo ĐLBT KL, có: m_A + m_O2 = m_CO2 + m_H2O

⇒ m_O2 = 0,8.44 + 18 - 11,6 = 41,6 (g)

\(\Rightarrow n_{O_2}=\dfrac{41,6}{32}=1,3\left(mol\right)\Rightarrow V_{O_2}=1,3.22,4=29,12\left(l\right)\)

a, \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

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

\(n_{Fe_3O_4}=\dfrac{17,4}{232}=0,075\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,075}{1}>\dfrac{0,2}{4}\), ta được Fe₃O₄ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{Fe_3O_4\left(pư\right)}=\dfrac{1}{4}n_{H_2}=0,05\left(mol\right)\\n_{Fe}=\dfrac{3}{4}n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{Fe_3O_4\left(dư\right)}=0,075-0,05=0,025\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4\left(dư\right)}+m_{Fe}=0,025.232+0,15.56=14,2\left(g\right)\)

Số mol của khí H₂;

n_H2 = \(\dfrac{V}{22.4}\) = \(\dfrac{4.48}{22.4}\) = 0.2 mol

Số mol của 17,4g Fe₃O_4:

n_Fe3O4 = \(\dfrac{m}{M}\) = \(\dfrac{17.4}{232}\) = 0,075 mol

PTHH: 2H₂ + Fe₃O₄ \(\rightarrow\)  3Fe + 4H₂O

Tỉ lệ :     2   :      1        :      3    :   4

Mol:     0,2      0.075   \(\rightarrow\) 0.225        

b.Khối lượng của Fe là

m_Fe = n.M = 0.225 . 56 = 12.6G