Đặt \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
Theo đề: \(m_{hh}=36\left(g\right)\)
\(\Rightarrow m_{Mg}+m_{Fe}=36\\ \Rightarrow24x+56y=36\left(1\right)\)
\(PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\\ \left(mol\right)....x\rightarrow...0,5x.....x\\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ \left(mol\right)....y\rightarrow...\dfrac{2}{3}y....\dfrac{1}{3}y\)
Theo đề: \(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
\(\Rightarrow0,5x+\dfrac{2}{3}y=0,6\left(2\right)\)
\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}24x+56y=36\\0,5x+\dfrac{2}{3}y=0,6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0,8\\y=0,3\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,8.24=19,2\left(g\right)\\m_{Fe}=0,3.56=16,8\left(g\right)\end{matrix}\right.\\ m_r=m_{MgO}+m_{Fe_3O_4}=0,8.40+\dfrac{1}{3}.0,3.232=55,2\left(g\right)\)
PTHH: C+O2→CO20,3mol:0,3mol→0,3molC+O2→CO20,3mol:0,3mol→0,3mol
S+O2→SO20,2mol:0,2mol→0,2molS+O2→SO20,2mol:0,2mol→0,2mol
mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)
VO2=(0,3+0,2)22,4=11,2(l)VO2=(0,3+0,2)22,4=11,2(l)
mhh=mCO2+mSO2=0,3.44+0,2.64=26(g)