a)
\(Mg+\dfrac{1}{2}O_2\underrightarrow{t^o}MgO\)
0,2<--0,1<--0,2
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(m_{Mg}=0,2.24=4,8\left(g\right)\\n_{O_2}=\dfrac{1}{2}.n_{MgO}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.24,79=2,479\left(l\right)\\ n_{O_2}=\dfrac{m_{MgO}-m_{Mg}}{32}=\dfrac{8-4,8}{32}=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.24,79=2,479\left(l\right)\)
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