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nS = 3,2/32 = 0,1 (mol)
nO2 = 8,96/22,4 = 0,4 (mol)
PTHH: S + O2 -> (t°) SO2
Mol: 0,1 ---> 0,1
nO2 (Al) = 0,4 - 0,1 = 0,3 (mol)
PTHH: 4Al + 3O2 -> (t°) 2Al2O3
Mol: 0,4 <--- 0,3
mAl = 0,4 . 27 = 10,8 (g)
mhh = 10,8 + 3,2 = 14 (g)
\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH:
2CO + O2 --to--> 2CO2
0,2<---0,1<--------0,2
2H2 + O2 --to--> 2H2O
0,4<--0,2<-------0,2
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,2}{0,2+0,4}.100\%=33,33\%\\\%V_{H_2}=100\%-33,33\%=66,67\%\end{matrix}\right.\)
\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: 2CO + O2 --to--> 2CO2
0,3<--0,15<------0,3
2H2 + O2 --to--> 2H2O
0,1<--0,05
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CO}=\%n_{CO}=\dfrac{0,3}{0,3+0,1}.100\%=75\%\\\%V_{H_2}=100\%-75\%=25\%\end{matrix}\right.\)
a/ \(2CO\left(0,2\right)+O_2\left(0,1\right)\rightarrow2CO_2\left(0,2\right)\)
\(2H_2\left(0,1\right)+O_2\left(0,05\right)\rightarrow2H_2O\left(0,1\right)\)
\(n_{H_2O}=\frac{1,8}{18}=0,1\)
\(n_{O_2}=\frac{3,36}{22,4}=0,15\)
Số mol O2 phản ứng ở phản ứng đầu là: \(0,15-0,05=0,1\)
\(\Rightarrow m_{CO_2}=0,2.44=8,8\)
b/ \(m_{CO}=0,2.28=5,6\)
\(m_{H_2}=0,1.2=0,2\)
c/ \(\%CO=\frac{0,2}{0,3}.100\%=66,67\%\)
\(\Rightarrow\%H_2=100\%-66,67\%=33,33\%\)
a)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
0,1<-0,05<-------0,1
2CO + O2 --to--> 2CO2
0,2<--0,1-------->0,2
=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\V_{CO}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)
b) \(m_{CO_2}=0,2.44=8,8\left(g\right)\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(n_{H_2O}=\dfrac{1,8}{18}=0,1mol\)
\(2CO+O_2\rightarrow2CO_2\)
a 0,5a a
\(2H_2+O_2\rightarrow2H_2O\)
0,1 0,05 \(\leftarrow\) 0,1
\(\Sigma n_{O_2}=0,5a+0,05=0,15\)
\(\Rightarrow a=n_{O_2\left(CO\right)}=0,2mol\)
\(V_{CO}=2\cdot0,2\cdot22,4=8,96l\)
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{CO_2}=0,2\cdot44=8,8g\)
\(n_{O_2}=\dfrac{3,36}{22,4}=0,15mol\)
Gọi \(\left\{{}\begin{matrix}n_C=x\\n_S=y\end{matrix}\right.\)
\(C+O_2\rightarrow\left(t^o\right)CO_2\)
x x ( mol )
\(S+O_2\rightarrow\left(t^o\right)SO_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}12x+32y=2,8\\x+y=0,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
\(\rightarrow m_C=0,1.12=1,2g\)
\(\rightarrow m_S=0,05.32=1,6g\)
\(\rightarrow\left\{{}\begin{matrix}\%m_C=\dfrac{1,2}{2,8}.100=42,85\%\\\%m_S=100\%-42,85\%=57,15\%\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_C=a\left(mol\right)\\n_S=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)
PTHH:
C + O2 --to--> CO2
a--->a
S + O2 --to--> SO2
b--->b
=> \(\left\{{}\begin{matrix}12a+32b=2,8\\a+b=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,05\left(mol\right)\end{matrix}\right.\left(TM\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_C=0,1.12=1,2\left(g\right)\\m_S=0,05.32=1,6\left(g\right)\end{matrix}\right.\)