Chương IV. Hiđrocacbon. Nhiên liệu - Hỏi đáp

M_C12H22O11 = 12.12+22+ 16.11 = 342

%C = \(\dfrac{12.12}{342}\) .100% \(\approx\) 42,1 %

%H = \(\dfrac{22}{342}.100\%\) \(\approx\) 6,43 %

%O = 100% - 42,1% - 6,43% = 51,47%

Thành phần phần trăm của nguyên tố C trong \(C_{12}H_{22}O_{11}\):

\(\%_C=\dfrac{m_C}{M_{C_{12}H_{22}O_{11}}}.100\%=\dfrac{12.12}{\left(12.12+22+11.16\right)}.100\%\approx42,11\%\)

\(\%_H=\dfrac{m_H}{M_{C_{12}H_{22}O_{11}}}.100\%=\dfrac{22}{\left(12.12+22+11.16\right)}.100\%\approx6,43\%\)

\(\%_O=100\%-\%_C-\%_H=100\%-42,11\%-6,43\%=51,46\%\)

\(n_{Fe}=\frac{m}{56}\left(mol\right)\)

\(n_{O_2\left(pư\right)}=\frac{12-m}{32}\left(mol\right)\)

\(n_{NO}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

Quá trình nhận-nhường e:

\(Fe^0-3e\rightarrow Fe^{+3}\)

\(\frac{m}{56}\)--->\(\frac{3m}{56}\)__________(mol)

\(O_2^0+4e\rightarrow2O^{-2}\)

\(\frac{12-m}{32}\)-> \(\frac{12-m}{8}\)________(mol)

\(NO_3^-+4H^++3e\rightarrow NO+2H_2O\)

______________0,3<----0,1____________(mol)

Áp dụng ĐLBT e: \(\frac{3m}{56}+\frac{12-m}{8}=0,3=>m=16,8\left(g\right)\)

Ta có pthh :

CaCO₃+ 2HCl---> CaCl₂ + H₂O + CO₂ ( 1)

(0,25) (0,25)

2Al + 3H₂SO₄---->Al₂(SO₄)₃ + 3H₂ (2)

( \(\frac{m}{27}\) ) ( \(\frac{3.m}{2.27}\))

nCaCO₃ =\(\frac{25}{100}\) = 0,25 mol ; nAl =\(\frac{m}{27}\)mol

Đặt mddHCl =mdd H₂SO₄= a g

m dd trong cốc A sau pư = 25 + a - 0,25.44 =(14 +a )g

mdd trong cốc B sau pư = m + a - \(\frac{3.m}{2.27}\) . 2

= ( m + a -\(\frac{m}{9}\)) g

Vì sau khi p.ư cân thăng bằng => 14 +a = m + a -\(\frac{m}{9}\)

<=> 14 = m - \(\frac{m}{9}\)

===> m = 15,75

Gọi \(m_{HCl}=x\left(g\right)\)

=> \(m_{H_2SO_4}=x\left(g\right)\)

\(n_{CaCO_3}=\frac{25}{100}=0,25\left(g\right)\)

\(n_{Al}=\frac{a}{27}\left(mol\right)\)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

_______0,25------------------------------->0,25_________(mol)

=> m_A (sau pư) = x + 25 - 0,25.44 = x + 14 (g)

PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

_______\(\frac{a}{27}\)-------------------------------------->\(\frac{a}{18}\)____(mol)

=> m_B (sau pư) = x + a - \(2.\frac{a}{18}\) = x + \(\frac{8}{9}a\) (mol)

Do sau pư, cân thăng bằng

=> \(x+14=x+\frac{8}{9}a\)

=> a = 15,75 (g)

C3H8 đó bạn

Ai giải hộ tui với

a,\(CH_4+2O_2\rightarrow CO_2+2H_2O\)

\(C_2H_4+3O_2\rightarrow2CO_2+2H_2O\)

\(n_{hh}=\frac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{CO2}=\frac{7,84}{22,4}=0,35\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{CH4}:a\left(mol\right)\\n_{C2H4}:b\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=0,25\\a+2b=0,35\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\end{matrix}\right.\)

\(V_{CH4}=0,15.22,4=3,36\left(l\right)\)

\(V_{C2H2}=0,1.22,4=2,24\left(l\right)\)

b,\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

\(n_{CaCO3}=n_{CO2}=0,35\left(mol\right)\)

\(\Rightarrow m_{CaCO3}-m_{CO2}=0,35.100-0,35.44=19,6\left(g\right)\)

a, PT: \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{Na_2O}=\frac{15,5}{62}=0,25\left(mol\right)\)

Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\frac{0,5}{0,5}=1M\)

b, PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Theo PT: \(n_{H_2SO_4}=\frac{1}{2}n_{NaOH}=0,25\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\frac{24,5.100\%}{20\%}=122,5\left(g\right)\)

Bạn tham khảo nhé!

mọi người giải theo cách đặt ẩn a và giải theo phương trình a được ko ạ e ko bít làm mấy cái tắt do e mới học lớp 9 ò

\(m_{CaCO_3}\) (thực tế) = \(\frac{100.90}{100}=90\left(g\right)\)

\(m_{CaCO_3\left(pư\right)}=\frac{90.85}{100}=76,5\left(g\right)\)

=> \(n_{CaCO_3}=\frac{76,5}{100}=0,765\left(mol\right)\)

PTHH: \(CaCO_3\underrightarrow{t^o}CaO+CO_2\uparrow\)

_______0,765------------->0,765_____(mol)

=> \(m_{CO_2}=0,765.44-33,66\left(g\right)\)