\(n_{S\left(đb\right)}=\frac{50}{32}=1,5625\left(mol\right)\)
\(n_{SO_2\left(đb\right)}=\frac{96}{64}=1,5\left(mol\right)\)
\(S+O_2\rightarrow SO_2\)
\(\frac{n_{S\left(đb\right)}}{n_{S\left(PTHH\right)}}\frac{n_{SO_2\left(đb\right)}}{n_{SO_2\left(PTHH\right)}}\)
\(\Rightarrow\frac{1,5625}{1}>\frac{1,5}{1}\)
\(\Rightarrow\) S dư, SO2 hết
Theo PTHH: \(n_S=n_{SO_2}=1,5\left(mol\right)\)
\(n_{O_2}=n_{SO_2}=1,5\left(mol\right)\)
\(\Rightarrow n_{S\left(dư\right)}=n_{S\left(đb\right)}-n_{S\left(PTHH\right)}=1,5625-1,5=0,0625\left(mol\right)\)
\(\Rightarrow m_{S\left(dư\right)}=0,0625.32=2\left(g\right)\)
\(\Rightarrow m_{O_2}=1,5.32=48\left(g\right)\)
nS(đb)=50\32=1,5625(mol)
nSO2=96\64=1,5(mol)
S+O2→SO2
(nS(đb)\nS(PTHH)(nSO2(đb)\nSO2(PT))
⇒1,5625\1>1,5\1
⇒⇒ S dư, SO2 hết
Theo PT: nS=nSO2=1,5(mol)
nO2=nSO2=1,5(mol)
⇒nS(dư)=nS(đb)−nS(PTHH)=1,5625−1,5=0,0625(mol)⇒nS(dư)=nS(đb)
⇒mS(dư)=0,0625.32=2(g)
⇒mO2=1,5.32=48(g)
