2Zn + O2 \(\underrightarrow{to}\) 2ZnO (1)
2Mg + O2 \(\underrightarrow{to}\) 2MgO (2)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
a) Theo PT1: \(n_{O_2}=\dfrac{1}{2}n_{Zn}=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
Theo pT2: \(n_{O_2}=\dfrac{1}{2}n_{Mg}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)\)
\(\Rightarrow\Sigma n_{O_2}=0,1+0,05=0,15\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,15\times22,4=3,36\left(l\right)\)
b) Theo PT1: \(n_{ZnO}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow m_{ZnO}=0,2\times81=16,2\left(g\right)\)
Theo pT2: \(n_{MgO}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,1\times4=4\left(g\right)\)
\(\Rightarrow m_{hhoxit}=16,2+4=20,2\left(g\right)\)
\(\%m_{ZnO}=\dfrac{16,2}{20,2}\times100\%=80,2\%\)
\(\%m_{MgO}=\dfrac{4}{20,2}\times100\%=19,8\%\)
nZn=m/M=13/65=0.2 mol
nMg=m/M=2.4/24=0.1 mol
2Zn + O2 -> 2ZnO
nZn=0.2 => nO2=0.1 mol nZnO=0.2 mol
2Mg + O2 ->2MgO
nMg=0.1 mol=> nO2= 0.05 mol nMgO= 0.1 mol
nO2 cần dùng= 0.1 + 0.05= 0.15 mol
a, VO2 cần dùng= n . 22.4= 3.36l
b, Oxit thu được gồm ZnO và MgO
mZnO= n.M= 0.2 . 81=16.2g
mMgO= n.M= 0.1 . 40=4g
m hỗn hợp oxit= mZnO+ mMgO= 16.2 + 4= 20.2g
%ZnO=(16.2 . 100%)/20.2 = 80.2%
%MgO= 100%-80.2%=19.8%
