a) \(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 2Mg + O2 --to--> 2MgO
0,8<--0,4-------->0,8
=> mMg = 0,8.24 = 19,2(g)
b)
C1: mMgO = 0,8.40 = 32(g)
C2: Theo ĐLBTKL: mMg + mO2 = mMgO
=> mMgO = \(19,2+0,4.32=32\left(g\right)\)
a: \(2Mg+O_2\rightarrow2MgO\)
\(n_{O_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(\Leftrightarrow n_{Mg}=0.8\left(mol\right)\)
\(m_{Mg}=0.8\cdot24=19.2\left(g\right)\)
b: \(m_{MgCl_2}=0.8\cdot40=32\left(g\right)\)
2Mg+O2-to>2MgO
0,8----0,4 -----0,8mol
n O2=\(\dfrac{8,96}{22,4}\)=0,1 mol
=>m Mg =0,8.24=19,2g
=>mMgO=0,8.56=44,8g
