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For the given reaction CH_(4(g))+2O_(2(g)) rarr CO_(2(g)) + 2H_(2)O_((l)) If 64 g of O_(2)is used, then 44 g ofCO_(2) is formed. 8 g of CH_(4)reacts to form 36 g of product. 22 g of CO_(2)is formed from 3.011xx10^(23)molecules of CH_(4). At STP, if 22.4 litres of O_(2(g)) is used, then 11.2 litres ofCO_(2) is formed. Which of the above statements are correct ? |
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Answer» `2, 3, 4` (1) Reaction : `CH_(4(g)) + 2O_(2(g)) rarr CO_(2(g)) +2H_(2)O_((l))` `64 g O_(2)` produces 44 gram `CO_(2)` `:.` 64 gram produces `:.` (1) is CORRECT (2) Reaction : `CH_(4(g))+2O_(2(g)) rarr CO_(2(g))+2H_(2)O_((l))` `:.` 16 gram `CH_(4)` react to form `(44+36)` `=80` gram product `(80xx8)/(16)` gram product `=40` gram product `:.` (2) is not correct (3)Reaction : `CH_(4(g))+2O_(2(g)) rarr CO_(2(g)) + 2H_(2)O_((l))` `:.` 1 mole `CH_(4)` form 1 mole `CO_(2)` gas `:.6.022 xx 10^(23)` molecules of `CH_(4)` form 44 g `CO_(2)`gas and `:.3.011xx10^(23)` molecules of `CH_(4)` form `=(3.011xx10^(23)xx44)/(6.022xx10^(23))=22g` of `CO_(2)` (3) is correct (4) Reaction : `CH_(4(g))+2O_(2(g)) rarr CO_(2(g))+2H_(2)O_((l))` `:.` At STP USED `2 (22.4L) O_(2)` form `22.4 L CO_(2)` `:.` At STP used `22.4L O_(2)` form `=(22.4Lxx22.4L)/(2xx22.4 L) CO_(2) = 11.2 ` of `CO_(2)` (4) is correct Thus, (1) , (3) and (4) are correct statements |
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