\(2Mg+O2-->2MgO\)
\(n_{MgO}=n_{Mg}=0,1\left(mol\right)\)
\(m_{MgO}=0,1.40=4\left(g\right)\)
nMg=0,1 mol
2Mg+O2-->2MgO
0,1-------------0,2 mol
=> mMgO=0,2,24=4,8g
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nMg=0,1 mol
2Mg+O2-->2MgO
0,1-------------0,2 mol
=> mMgO=0,2,24=4,8g