Gọi CTHH của hợp chất gồm C và S là CxSy
(x,y \(\in N^{star}\))
Ta có:
\(x:y=\dfrac{m_C}{M_C}:\dfrac{m_S}{M_S}=\dfrac{2,4}{12}:\dfrac{9,6}{32}=0,2:0,3=\dfrac{2}{3}\)
Chọn x = 2, y = 3
Vậy CTHH là C2S3
PTHH: C2S3 + 5O2 ------> 2CO2 + 3SO2
a.
\(\%m_C=\dfrac{M_C\cdot2}{M_{C_2S_3}}\cdot100\%=\dfrac{12\cdot2}{120}\cdot100\%=20\%\)
\(\Rightarrow\%m_S=100\%-20\%=80\%\)
b.
\(n_{C_2S_3}=\dfrac{m_{C_2S_3}}{M_{C_2S_3}}=\dfrac{2,4+9,6}{120}=0,2\left(mol\right)\)
PTHH: C2S3 + 5O2 ------> 2CO2 + 3SO2
1 mol 5 mol
0,2 mol x mol
\(\Rightarrow x=\dfrac{0,2\cdot5}{1}=1\left(mol\right)\)
\(\Rightarrow V_{O_2}=n_{O_2}\cdot22,4=1\cdot22,4=22,4\left(l\right)\)
c.
PTHH: C2S3 + 5O2 ------> 2CO2 + 3SO2
1 mol 2 mol 3 mol
0,2 mol x mol y mol
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{0,2\cdot2}{1}=0,4\left(mol\right)\\y=\dfrac{0,2\cdot3}{1}=0,6\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{CO_2}+V_{SO_2}=n_{CO_2}\cdot22,4+n_{SO_2}\cdot22,4=22,4\cdot\left(0,4+0,6\right)=22,4\cdot1=22,4\left(l\right)\)
