Ta có:
\(m_{Al}:m_{Mg}=21:16\\ \rightarrow\dfrac{m_{Al}}{21}=\dfrac{m_{Mg}}{16}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{m_{Al}}{21}=\dfrac{m_{Mg}}{16}=\dfrac{m_{Al}+m_{Mg}}{21+16}=\dfrac{6,66}{37}=0,18\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,18.21=3,78\left(g\right)\\m_{Mg}=0,18.16=2,88\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{3,78}{27}=0,14\left(mol\right)\\n_{Mg}=\dfrac{2,88}{24}=0,12\left(mol\right)\end{matrix}\right.\)
PTHH:
4Al + 3O2 --to--> 2Al2O3
0,14---------------->0,07
2Mg + O2 --to--> 2MgO
0,12--------------->0,12
=> m = 0,07.102 + 0,12.40 = 11,94 (g)
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