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༺ミ𝒮σɱєσиє...彡༻ · Accepted Answer

a)ta có $n_{N_2}=\dfrac{4,2}{28}=0,15\left(mol\right)$$\Rightarrow V_{N_2}=0,15.22,4=3,36l$ta có $n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)$$\Rightarrow V_{O_2}=0,1.22,4=2,24l$b) ta có $n_{SO_2}=\dfrac{32}{64}=0,5\left(mol\right)$$\Rightarrow V_{SO_2}=0,5.22,4=11,2l$ta có $n_{O_2}=\dfrac{32}{32}=1\left(mol\right)$$\Rightarrow V_{O2}=1.22,4=22,4l$c) ta có $n_{H_2}=\dfrac{2}{2}=1\left(mol\right)$$\Rightarrow V_{H_2}=1.22,4=22,4l$ta có $n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)$$\Rightarrow V_{O_2}=0,25.22,4=5,6l$d) ta có $n_{N_2}=\dfrac{8,4}{28}=0,3\left(mol\right)$$\Rightarrow V_{N_2}=0,3.22,4=6,72l$ta có $n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)$$\Rightarrow V_{N_2}=0,2.22,4=4,48l$Câu trả lời: a)ta có $n_{N_2}=\dfrac{4,2}{28}=0,15\left(mol\right)$$\Rightarrow V_{N_2}=0,15.22,4=3,36l$ta có $n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)$$\Rightarrow V_{O_2}=0,1.22,4=2,24l$b) ta có $n_{SO_2}=\dfrac{32}{64}=0,5\left(mol\right)$$\Rightarrow V_{SO_2}=0,5.22,4=11,2l$ta có $n_{O_2}=\dfrac{32}{32}=1\left(mol\right)$$\Rightarrow V_{O2}=1.22,4=22,4l$c) ta có $n_{H_2}=\dfrac{2}{2}=1\left(mol\right)$$\Rightarrow V_{H_2}=1.22,4=22,4l$ta có $n_{O_2}=\dfrac{8}{32}=0,25\left(mol\right)$$\Rightarrow V_{O_2}=0,25.22,4=5,6l$d) ta có $n_{N_2}=\dfrac{8,4}{28}=0,3\left(mol\right)$$\Rightarrow V_{N_2}=0,3.22,4=6,72l$ta có $n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)$$\Rightarrow V_{N_2}=0,2.22,4=4,48l$

Bài 18: Mol

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