Bài 9:
a, PT: \(2Zn+O_2\underrightarrow{t^o}2ZnO\)
\(4Al+3O_2\underrightarrow{t^{^o}}2Al_2O_3\)
b, Ta có: m tăng = mO2 = 6 (g)
\(\Rightarrow n_{O_2}=\dfrac{6}{36}=\dfrac{1}{6}\left(mol\right)\)
Gọi: nZn = nAl = x (mol)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Zn}+\dfrac{3}{4}n_{Al}=\dfrac{1}{2}x+\dfrac{3}{4}x=\dfrac{1}{6}\) \(\Rightarrow x=\dfrac{2}{15}\left(mol\right)\)
\(\Rightarrow m=m_{Zn}+m_{Al}=\dfrac{2}{15}.65+\dfrac{2}{15}.27\approx12,267\left(g\right)\)
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