Bài 1 : Sửa ZnSO thành ZnSO₄

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

Pt : \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Theo Pt : \(n_{Zn}=n_{H2SO4}=n_{ZnSO4}=n_{H2}=0,2\left(mol\right)\)

a) \(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

b) \(m_{ZnSO4}=0,2.161=32,2\left(g\right)\)

c) \(V_{H2\left(dkc\right)}=0,2.24,79=4,958\left(l\right)\)

Bài 3 : 

\(n_{O2}=\dfrac{9,6}{32}=0,3\left(mol\right)\)

\(2KClO_3\xrightarrow[]{t^o}2KCl+3O_2\)

  0,2<-----------0,2<----0,3

a) \(m_{KClO3}=0,2.122,5=24,5\left(g\right)\)

b) Cách 1 :  \(m_{KCl}=0,2.74,5=14,9\left(g\right)\)

    cách 2 : \(BTKl:m_{KClO3}=m_{KCl}+m_{O2}\)

                          \(\Rightarrow m_{KCl}=m_{KClO3}-m_{O2}=24,5-9,6=14,9\left(g\right)\)

Đổi \(200ml=0,2l\)

Số mol của 5,85g NaCl là:

\(n_{NaCl}=\dfrac{m_{NaCl}}{M_{NaCl}}=\dfrac{5,85}{58,5}=0,1\left(mol\right)\)

Nồng độ mol của \(5,85g\) \(NaCl\) là:

\(C_MNaCl=\dfrac{n}{V}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(n_{NO_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)

\(n_{CO_2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)

\(n_{N_2O}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)

\(n_{Cl_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

\(n_{SO_2}=\dfrac{0,115}{22,4}=\dfrac{23}{4480}\left(mol\right)\)

\(n_{H_2S}=\dfrac{0,225}{22,4}=\dfrac{9}{896}\left(mol\right)\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

Theo PT: \(n_{Fe_3O_4\left(LT\right)}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{30}\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4\left(LT\right)}=\dfrac{1}{30}.232=\dfrac{116}{15}\left(g\right)\)

\(\Rightarrow H=\dfrac{\dfrac{116}{15}}{11,6}.100\%\approx66,67\%\)

\(n_{Fe_3O_4}=\dfrac{1}{30}\left(mol\right)\) chứ không phải \(\dfrac{1}{3}\) em nhé.

Ta có: \(d_{H_2/kk}=\dfrac{M_{H_2}}{29}=\dfrac{2}{29}< 1\)

⇒ H₂ nhẹ hơn không khí nên bị không khí đẩy bay lên trên.

đề hỏi gì vậy bạn?

a/ tính klg chất dư sau phản úng 

.b/ tính thể tích h2đkc sinh ra

Ta có: \(n_{C_6H_5SO_2NNaCl}=\dfrac{0,25}{213,5}=\dfrac{1}{854}\left(mol\right)\)

\(\Rightarrow C_{M_{C_6H_5SO_2NNaCl}}=\dfrac{\dfrac{1}{854}}{25}\approx4,68.10^{-5}\left(M\right)\)

→ Đáp án: A

Đáp án: A

a, \(2Cu\left(NO_3\right)_2\underrightarrow{t^o}2CuO+4NO_2+O_2\)

b, \(n_{Cu\left(NO_3\right)_2}=\dfrac{28,2}{188}=0,15\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{CuO}=n_{Cu\left(NO_3\right)_2}=0,15\left(mol\right)\\n_{O_2}=\dfrac{1}{2}n_{Cu\left(NO_3\right)_2}=0,075\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)

\(V_{O_2}=0,075.24,79=1,85925\left(l\right)\)

c, Ta có: \(n_{NO_2}+n_{O_2}=\dfrac{6,1975}{24,79}=0,25\left(mol\right)\)

Gọi: n_O2 = x (mol)

Theo PT: \(n_{NO_2}=4n_{O_2}=4x\left(mol\right)\)

⇒ 4x + x = 0,25 ⇒ x = 0,05 (mol)

Theo PT: \(n_{Cu\left(NO_3\right)_2\left(LT\right)}=2n_{O_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Cu\left(NO_3\right)_2\left(LT\right)}=0,1.188=18,8\left(g\right)\)

Mà: H = 80% \(\Rightarrow m_{Cu\left(NO_3\right)_2\left(TT\right)}=\dfrac{18,8}{80\%}=23,5\left(g\right)\)