e lạy mấy a, mấy chị vào giải dùm e bài này với ạ!!

trời đậu!!

\(1,2H_2+O_2\underrightarrow{t}2H_2O\)

\(2Mg+O_2\underrightarrow{t}2MgO\)

\(2Cu+O_2\underrightarrow{t}2CuO\)

\(S+O_2\underrightarrow{t}SO_2\)

\(4Al+3O_2\underrightarrow{t}2Al_2O_3\)

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

\(4P+5O_2\underrightarrow{t}2P_2O_5\)

\(2,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(a,n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)

\(b,n_C=0,3\left(mol\right)\Rightarrow n_{CO_2}=0,3\left(mol\right)\Rightarrow m_{CO_2}=13,2\left(g\right)\)

c, Vì\(\frac{0,3}{1}>\frac{0,2}{1}\)nên C phản ửng dư, O₂ phản ứng hết, Bài toán tính theo O₂

\(n_{O_2}=0,2\left(mol\right)\Rightarrow n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CO_2}=8,8\left(g\right)\)

\(3,PTHH:CH_4+2O_2\underrightarrow{t}CO_2+2H_2O\)

\(C_2H_2+\frac{5}{2}O_2\underrightarrow{t}2CO_2+H_2O\)

\(C_2H_6O+3O_2\underrightarrow{t}2CO_2+3H_2O\)

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

\(n_P=1,5\left(mol\right)\Rightarrow n_{O_2}=1,2\left(mol\right)\Rightarrow m_{O_2}=38,4\left(g\right)\)

\(b,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(n_C=2,5\left(mol\right)\Rightarrow n_{O_2}=2,5\left(mol\right)\Rightarrow m_{O_2}=80\left(g\right)\)

\(c,PTHH:4Al+3O_2\underrightarrow{t}2Al_2O_3\)

\(n_{Al}=2,5\left(mol\right)\Rightarrow n_{O_2}=1,875\left(mol\right)\Rightarrow m_{O_2}=60\left(g\right)\)

\(d,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)

\(TH_1:\left(đktc\right)n_{H_2}=1,5\left(mol\right)\Rightarrow n_{O_2}=0,75\left(mol\right)\Rightarrow m_{O_2}=24\left(g\right)\)

\(TH_2:\left(đkt\right)n_{H_2}=1,4\left(mol\right)\Rightarrow n_{O_2}=0,7\left(mol\right)\Rightarrow m_{O_2}=22,4\left(g\right)\)

\(5,PTHH:S+O_2\underrightarrow{t}SO_2\)

\(n_{O_2}=0,46875\left(mol\right)\)

\(n_{SO_2}=0,3\left(mol\right)\)

Vì\(0,46875>0,3\left(n_{O_2}>n_{SO_2}\right)\)nên S phản ứng hết, bài toán tính theo S.

\(a,\Rightarrow n_S=n_{SO_2}=0,3\left(mol\right)\Rightarrow m_S=9,6\left(g\right)\)

\(n_{O_2}\left(dư\right)=0,16875\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=5,4\left(g\right)\)

\(6,a,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_C=1,5\left(mol\right)\Rightarrow m_C=18\left(g\right)\)

\(b,PTHH:2H_2+O_2\underrightarrow{t}2H_2O\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_{H_2}=0,75\left(mol\right)\Rightarrow m_{H_2}=1,5\left(g\right)\)

\(c,PTHH:S+O_2\underrightarrow{t}SO_2\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_S=1,5\left(mol\right)\Rightarrow m_S=48\left(g\right)\)

\(d,PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)

\(n_{O_2}=1,5\left(mol\right)\Rightarrow n_P=1,2\left(mol\right)\Rightarrow m_P=37,2\left(g\right)\)

\(7,n_{O_2}=5\left(mol\right)\Rightarrow V_{O_2}=112\left(l\right)\left(đktc\right)\);\(V_{O_2}=120\left(l\right)\left(đkt\right)\)

\(8,PTHH:C+O_2\underrightarrow{t}CO_2\)

\(m_C=0,96\left(kg\right)\Rightarrow n_C=0,08\left(kmol\right)=80\left(mol\right)\Rightarrow n_{O_2}=80\left(mol\right)\Rightarrow V_{O_2}=1792\left(l\right)\)

\(9,n_p=0,2\left(mol\right);n_{O_2}=0,3\left(mol\right)\)

\(PTHH:4P+5O_2\underrightarrow{t}2P_2O_5\)

Vì\(\frac{0,2}{4}< \frac{0,3}{5}\)nên P hết O2 dư, bài toán tính theo P.

\(a,n_{O_2}\left(dư\right)=0,05\left(mol\right)\Rightarrow m_{O_2}\left(dư\right)=1,6\left(g\right)\)

\(b,n_{P_2O_5}=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=14,2\left(g\right)\)

đủ cả 9 câu bạn nhé,

Gọi số mol C, S là a, b

=> 12a + 32b = 7,68

PTHH: C + O₂ --t^o--> CO₂

_____a--------------->a

S + O₂ --t^o--> SO₂

b--------------->b

=> a + b = \(\dfrac{9,856}{22,4}=0,44\)

=> a = 0,32; b = 0,12

=> \(\left\{{}\begin{matrix}\%C=\dfrac{0,32.12}{7,68}.100\%=50\%\\\%S=\dfrac{0,12.32}{7,68}.100\%=50\%\end{matrix}\right.\)

Hóa mà s lại Toán ??

a: \(C+O_2\rightarrow CO_2\)(ĐK: t độ)

x         x          x

\(S+O_2\rightarrow SO_2\)(ĐK: t độ)

y       y         y

b: \(n_{O_2}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

Theo đề, ta có hệ:

12x+32y=10 và x+y=0,5

=>x=0,3 và y=0,2

\(m_C=0.3\cdot12=3.6\left(g\right)\)

\(m_S=0.2\cdot32=6.4\left(g\right)\)

c: \(n_{CO_2}=n_C=0.3\left(mol\right)\)

\(n_{SO_2}=n_S=0.2\left(mol\right)\)

\(V_{khí}=22.4\left(0.3+0.2\right)=11.2\left(lít\right)\)

\(n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH :

\(C+O_2\rightarrow\left(t^o\right)CO_2\)

x       x               x 

\(S+O_2\rightarrow\left(t^o\right)SO_2\)

y      y                y

Gọi n C = x 

     n S = y (mol)

Ta có hệ PT :

\(\left\{{}\begin{matrix}12x+32y=10\\x+y=0,5\end{matrix}\right.\)

\(\rightarrow x=0,3;y=0,2\)

\(m_C=0,3.12=3,6\left(g\right)\)

\(m_S=0,2.32=6,4\left(g\right)\)

\(c,V_{hhk}=\left(0,3+0,2\right).22,4=11,2\left(l\right)\)

Ta có: \(n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\)

a. PTHH: S + O₂ ---t^o---> SO₂

Theo PT: \(n_{SO_2}=n_S=0,2\left(mol\right)\)

=> \(m_{SO_2}=0,2.64=12,8\left(g\right)\)

b. Theo PT: \(n_{O_2}=n_S=0,2\left(mol\right)\)

=> \(m_{O_2}=0,2.32=6,4\left(g\right)\)

a)S+O2-------->SO2

b)n S=6,4/32=0,2(mol)

Theo pthh

n SO2 =n S=0,2(mol)

V SO2=0,2.22,4=4,48(mol)

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a, \(n_S=\dfrac{6,4}{32}=0,2\left(mol\right)\)

PTHH: S + O₂ ----t^o----> SO₂

Mol:    0,2   0,2                 0,2

b, \(m_{SO_2}=0,2.64=12,8\left(g\right)\)

c, \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)