\(n_S=\frac{50}{32}=1,5625\left(mol\right)\)
\(n_{SO2}=\frac{96}{64}=1,5\left(mol\right)\)
a. \(PTHH:S+O_2\rightarrow SO_2\)
b. Theo PT ta có:
\(n_{O2}=n_{SO2}=1,5\left(mol\right)\)
\(\Rightarrow m_{O2}=105.32=48\left(g\right)\)
\(\Rightarrow V_{O2}=1,5.22,4=33,6\left(l\right)\)
Theo ĐLBTKL ta có:
\(m_S+m_{O2}=m_{SO2}\)
\(m_S=m_{SO2}-m_{O2}=96-48=48\left(g\right)\)
\(n_{S\left(dư\right)}=50-48=2\left(g\right)\)