\(n_C=\dfrac{24}{12}=2\left(mol\right)\)

Pt : \(C+O_2\underrightarrow{t_o}CO_2|\)

       1      1       1

       2     2

\(n_{O2}=\dfrac{2.1}{1}=2\left(mol\right)\)

\(V_{O2\left(dktc\right)}2.22,4=44,8\left(l\right)\)

 Chúc bạn học tốt

\(n_C=\dfrac{14,4}{44}=\dfrac{18}{55}\left(mol\right)\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ n_{O_2}=n_C=n_{CO_2}=\dfrac{18}{55}\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=\dfrac{18}{55}.22,4=\dfrac{2016}{275}\left(lít\right)\\ b,V_{kk}=\dfrac{100}{21}.\dfrac{2016}{275}=\dfrac{381}{11}\left(lít\right)\\ c,2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3\left(LT\right)}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.\dfrac{18}{55}=\dfrac{12}{55}\left(mol\right)\\ \Rightarrow n_{KClO_3\left(TT\right)}=120\%.\dfrac{12}{55}=\dfrac{72}{275}\left(mol\right)\\ \Rightarrow m_{KClO_3}=122,5.\dfrac{72}{275}=\dfrac{1764}{55}\left(g\right)\)

giúp mik với trong tối nay plsssss

\(n_C=\dfrac{24}{12}=2\left(mol\right)\\ C+O_2\underrightarrow{^{to}}CO_2\\ n_{O_2}=n_C=2\left(mol\right)\\ V_{O_2\left(đktc\right)}=2.22,4=4,48\left(l\right)\)

tham khảo tại đây nka bn: https://hoc24.vn/cau-hoi/tinh-the-tich-khi-oxi-o-dktc-can-dung-de-dot-chay-het-24g-than-cacbon.1857970648802

C + \(O_2\) ➙ \(CO_2\)

Số mol của Cacbon là: \(n_C\) = \(\dfrac{24}{12}\) = 2 (mol)

Số mol của Oxi là: \(n_{O_2}\) = \(n_C\) = 2 (mol)

Thể tích Oxi cần dùng là: V = \(n_{O_2}\) x 22,4 = 44,8 (l)

\(TH1:n_C=\dfrac{m}{M}=\dfrac{24}{12}=2\left(mol\right)\\ PTHH:O_2+C\underrightarrow{t^o}CO_2\\ Theo.PTHH\Rightarrow n_{O_2}=n_C=2\left(mol\right)\\ V_{O_2\left(dktc\right)}=n.22,4=2.22,4=44,8\left(l\right)\)

\(TH2:n_C=\dfrac{m}{M}=\dfrac{24}{12}=2\left(mol\right)\\ PTHH:O_2+2C\underrightarrow{t^o}2CO\\ Theo.PTHH:n_{O_2}=\dfrac{1}{2}n_C=\dfrac{1}{2}.2=1\left(mol\right)\\ V_{O_2\left(dktc\right)}=n.22,4=1.22,4=22,4\left(l\right)\)

C+O₂-to>CO₂

1,5--1,5 -----1,5mol

n C=\(\dfrac{18}{12}\)=1,5 mol

=>V_kk=1,5.22,4.5=168l

=>V_CO2=1,5.22,4=33,6l

nC = 18/12 = 1,5 (mol)

PTHH: C + O2 -> (t°) CO2

Mol: 1,5 ---> 1,5 ---> 1,5 

VO2 = 1,5 . 22,4 = 33,6 (l)

Vkk = 33,6 . 5 = 168 (l)

VCO2 = 1,5 . 22,4 = 33,6 (l)

Phương trình phản ứng khi đốt cháy cacbon:

V O 2 = n O 2 .22,4 = 1.22,4 = 22,4(l)

→ V k k = 5 V O 2  = 5.22,4 = 112(l)

\(n_{C_2H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{^{^{t^0}}}2CO_2+H_2O\)

\(0.05.......0.125........0.1\)

\(V_{CO_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(V_{kk}=5\cdot V_{O_2}=5\cdot0.125\cdot22.4=14\left(l\right)\)

→⎧⎪ ⎪⎨⎪ ⎪⎩nC=60012=50(mol)nS=832=0,25(mol)→{nC=60012=50(mol)nS=832=0,25(mol)

PTHH:

C + O₂ ---t^o----> CO₂

_{50.....50.......................................(mol)}

S + O₂ ----t^o---> SO₂

^{0,25...0,25 ............................................(mol)}

→ VO2VO2 _{(cần dùng)} = (50 + 0,25) . 22,4 = 1125,6 (l)

=>Vkk=1125.6.5=5628l

\(n_C=\dfrac{24}{12}=2\left(mol\right)\)

\(C+O_2\underrightarrow{^{t^0}}CO_2\)

\(2.......2\)

\(V_{kk}=5V_{O_2}=5\cdot2\cdot22.4=224\left(l\right)\)

Oxi chiếm 20% em nhé !

Oxi chiếm 20%.