Question

Answer 1

\(m_S=\dfrac{40.80}{100}=32\left(g\right)=>n_S=\dfrac{32}{32}=1\left(mol\right)\)

\(m_O=\dfrac{60.80}{100}=48\left(g\right)=>n_O=\dfrac{48}{16}=3\left(mol\right)\)

=> CTHH: SO₃