PTHH: \(2KMnO_4\underrightarrow{to}K_2MnO_4+MnO_2+O_2\\ 0,5mol\rightarrow0,25mol:0,25mol:0,25mol\)
\(n_{KMnO_4}=\dfrac{79}{158}=0,5\left(mol\right)\)
PTHH: \(S+O_2\underrightarrow{to}SO_2\\ 0,25mol:0,25mol\rightarrow0,25mol\)
\(n_S=\dfrac{128}{32}=4\left(mol\right)\)
Ta có: \(4>0,25\)
Vậy Oxi phản ứng hết, Lưu huỳnh phản ứng dư.
\(n_{Sdu}=4-0,25=3,75\left(mol\right)\)
\(m_{Sdu}=3,75.32=120\left(g\right)\)
\(m_{SO_2}=0,25.64=16\left(g\right)\)
\(m_{Sanpham}=m_{Sdu}+m_{SO_2}=120+16=136\left(g\right)\)
2KMnO4 \(\underrightarrow{to}\) K2MnO4 + MnO2 + O2 (1)
\(n_{KMnO_4}=\dfrac{79}{158}=0,5\left(mol\right)\)
Theo PT1: \(n_{O_2}=\dfrac{1}{2}n_{KMnO_4}=\dfrac{1}{2}\times0,5=0,25\left(mol\right)\)
\(n_S=\dfrac{128}{32}=4\left(mol\right)\)
PTHH: S + O2 \(\underrightarrow{to}\) SO2 (2)
Bđ : 4..........0,25...............(mol)
Pư. : 0,25.....0,25................(mol)
Sau pư: 3,75......0.....→.....0,25.(mol)
Sản phẩm sau phản ứng gồm: SO2 và S dư
\(m_{SO_2}=0,25\times64=16\left(g\right)\)
\(m_Sdư=3,75\times32=120\left(g\right)\)
\(\Rightarrow m_{sp}=m_{SO_2}+m_Sdư=16+120=136\left(g\right)\)
