Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\)
\(PTHH:2Mg+O_2\overset{t^o}{--->}2MgO\)
a. Ta thấy: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\)
Vậy oxi dư.
Theo PT: \(n_{O_{2_{PỨ}}}=\dfrac{1}{2}.n_{Mg}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)
\(\Rightarrow n_{O_{2_{dư}}}=0,2-0,1=0,1\left(mol\right)\)
b. Theo PT: \(n_{MgO}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)
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