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From 200 mg of CO(2), 10^(21) molecules ...

From `200 mg` of `CO_(2)`, `10^(21)` molecules are removed. How many grams and moles of `CO_(2)` are left?

Text Solution

Verified by Experts

`:' 6.023 xx 10^(23)` molecules of `CO_(2) = 44 g`
`:. 10^(21)` molecules of `CO_(2) = (44 xx 10^(21))/(6.023 xx 10^(23))`
`= 7.31 xx 10^(-2) = 73.1 mg`
`:. CO_(2)` left `= 200 - 73.1 = 126.9 mg`
Also, moles of `CO_(2) = ("Weight")/("Molecular Weight") = (126.9 xx 10^(-3))/(44)`
`= 2.88 xx 10^(-3)`
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