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From `200 mg` of `CO_(2), 10^(21)` molecules are removed. How many grams and moles of `CO_(2)` are left. |
Answer» `44 g of CO_(2) = 1 mol = 6.023 xx 10^(23)` molecules `:. 6.023 xx 10^(23)` molecules `= 44 g of CO_(2)` `10^(21)` molecules `= (40 xx 10^(21) xx 10^(3))/(6.023 xx 10^(23)) mg` Weight of `CO_(2)` left `= 200 - 73.05 = 126.9 mg` `(126.9)/(10^(3)) = 0.1269 g` `44 g of CO_(2) = 1 mol` `0.1269 g of CO_(2) = (1)/(44) xx 0.1269 = 0.0028 mol` |
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