(1)4.48L 2.24L (3分)(2) 3.92 L (3分)
(1)设NO和NO 2的体积分别为x L与yL,则x+y =22.4
由电子得失守恒:(x/22.4)×3 + (y/22.4)×1=(22.4/64)×2
解得:x=4.48L y=2.24L
(2)由方程式知:4NO+3O 2+2H 2O=4HNO 34NO 2+O 2+2H 2O=4HNO 3可求得氧气的体积为3.92 L
(1)4.48L 2.24L (3分)(2) 3.92 L (3分)
(1)设NO和NO 2的体积分别为x L与yL,则x+y =22.4
由电子得失守恒:(x/22.4)×3 + (y/22.4)×1=(22.4/64)×2
解得:x=4.48L y=2.24L
(2)由方程式知:4NO+3O 2+2H 2O=4HNO 34NO 2+O 2+2H 2O=4HNO 3可求得氧气的体积为3.92 L