Question

Answer 1

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right);n_{SO_2}=\dfrac{6,4}{64}=0,1\left(mol\right)\\ S+O_2\rightarrow\left(t^o\right)SO_2\\ Vì:\dfrac{0,1}{1}=\dfrac{0,1}{1}\Rightarrow P.ứ.hết\\ n_{O_2}=n_S=0,1\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

Answer 2

S+O₂->SO₂

Bảo toàn khối lượng :

mS+mO₂=mSO₂

m_O2=6,4-3,2=3,2g=>n _O2=\(\dfrac{3,2}{32}\)=0,1 mol

=>V_O2=0,1.22,4=2,24l

Answer 3

\(n_S=\dfrac{3,2}{32}=0,1\)

\(PTHH:S+O_2=SO_2\)

              1    :  1     :  1

            0,1  :  0,1   : 0,1

\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(g\right)\)

Vậy..........