a) PTHH: \(S+O_2\underrightarrow{t^o}SO_2\)
Ta có: \(n_{O_2}=n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\) \(\Rightarrow V_{O_2}=0,1\cdot22,4=2,24\left(l\right)\)
b) PTHH: \(C+O_2\underrightarrow{t^o}CO_2\)
Ta có: \(n_{O_2}=n_C=\dfrac{6}{12}=0,5\left(mol\right)\) \(\Rightarrow V_{O_2}=0,5\cdot22,4=11,2\left(l\right)\)
a/
Áp dụng công thức \(m=n.M=>n=\dfrac{m}{M}\)
\(=>n_S=\dfrac{m_S}{M_S}=\dfrac{3.2}{32}=0.1\left(mol\right)\)
PTHH
\(S+O_2\underrightarrow{t^o}SO_2\)
1 1
0.1 x
=>\(x=0.1\cdot1:1=0.1=n_{O_2}\)
\(=>V_{O_2}=0.1\cdot22,4=2.24\left(l\right)\)
a. PT: S + O2 ---> SO2.
Ta có: nS=3,2/32=0,1(mol)
Theo PT, ta có: nS=nO=0,1(mol)
=> VO=0,1.22,4=2,24(l)
=> Vkk=5.VO=5.2,24=11,2(l)
b. PT: C + O2 ---> CO2.
Ta có: nC=6/12=0,5(mol)
Theo PT, ta có: nC=nO=0,5(mol)
=> VO=0,5.22,4=11,2(l)
=> => Vkk=5.VO=5.11,2=56(l)