a) PTHH: 2Mg + O2 → 2MgO
Đây là phản ứng oxi hóa
b) \(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
Theo PT: \(n_{MgO}=n_{Mg}=0,5\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,5\times40=20\left(g\right)\)
c) Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Mg}=\dfrac{1}{2}\times0,5=0,25\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,25\times22,4=5,6\left(l\right)\)
nMg=0,5(mol)
2Mg + O2 -> 2MgO (1) (PƯ oxi hóa)
0,5 0,25 0,5
mMgO=0,5.40=20(g)
VO2=0,25.22,4=5,6(lít)
a) PTHH: 2Mg + O2 → 2MgO(phản ứng oxi hóa)
b) n\(_{Mg}\)=1224=0,5(mol)n\(_{Mg}\)=1224=0,5(mol)
Theo PT: n\(_{MgO}\)=n\(_{Mg}\)=0,5(mol)n\(_{MgO}\)=nMg=0,5(mol)
⇒m\(_{MgO}\)=0,5×40=20(g)⇒m\(_{Mg}\)O=0,5×40=20(g)
c) Theo PT: n\(_O\)\(_2\)=12n\(_{Mg}\)=12×0,5=0,25(mol)
n\(_O\)\(_2\)=12n\(_{Mg}\)=12×0,5=0,25(mol)
⇒V\(_O\)\(_2\)=0,25×22,4=5,6(l)⇒V\(_O\)\(_2\)=0,25×22,4=5,6(l)
