\(Ta.có:\dfrac{n_{Mg}}{n_{Zn}}=\dfrac{3}{1}\Rightarrow\dfrac{\dfrac{m_{Mg}}{M_{Mg}}}{\dfrac{m_{Zn}}{M_{Zn}}}=\dfrac{m_{Mg}}{24}.\dfrac{65}{m_{Zn}}=\dfrac{3}{1}\Rightarrow\dfrac{m_{Mg}}{m_{Zn}}=\dfrac{3}{1}:\dfrac{65}{24}=\dfrac{72}{24}\)
\(\Rightarrow n_{Mg}=13,7:\left(72+24\right).3=0,3\left(mol\right)\\ \Rightarrow n_{Zn}=\dfrac{n_{Mg}}{3}=\dfrac{0,3}{3}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=n.M=0,3.24=7,2\left(g\right)\\ \Rightarrow m_{Zn}=n.M=0,12.65=6,5\left(g\right)\)
\(a,PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\left(1\right)\\ PTHH:2Zn+O_2\underrightarrow{t^o}2ZnO\left(2\right)\)
\(Theo.PTHH\left(1\right):n_{O_2\left(1\right)}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,3=0,15\left(mol\right)\\ Theo.PTHH\left(2\right):n_{O_2\left(2\right)}=\dfrac{1}{2}.n_{Zn}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ n_{O_2\left(tổng\right)}=n_{O_2\left(1\right)}+n_{O_2\left(2\right)}=0,15+0,05=0,2\left(mol\right)\\ V_{O_2\left(tổng,đktc\right)}=n.22,4=0,2.22,4=4,48\left(l\right)\)
\(b,Theo.PTHH\left(1\right):n_{MgO}=n_{Mg}=0,3\left(mol\right)\\ m_{MgO}=n.M=0,3.40=12\left(g\right)\\ Theo.PTHH\left(2\right):n_{ZnO}=n_{Zn}=0,1\left(mol\right)\\ m_{ZnO}=n.M=0,1.81=8,1\left(g\right)\\ m=m_{hh}=m_{MgO}+m_{ZnO}=12+8,1=20,1\left(g\right)\)