\(a,n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ a,4K+O_2\rightarrow2K_2O\\ Vì:\dfrac{0,1}{4}< \dfrac{0,15}{1}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=0,15-\dfrac{1}{4}.0,1=0,125\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=0,125.32=4\left(g\right)\\ b,n_{K_2O}=\dfrac{2}{4}.n_K=\dfrac{1}{2}.0,1=0,05\left(mol\right)\\ \Rightarrow m_{K_2O}=94.0,05=4,7\left(g\right)\)
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