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

Explanation

Gram molecular mass of `CO_(2) = 44g`
Mass of `6.022 xx 10^(23)` molecules of `CO_(2) = 44 g`
Mass of `10^(21)` molecules of `CO_(2) = 44/(6.022 xx 10^(23)) xx 10^(21) = 0.073 g`
Mass of `CO_(2)` left `=(0.2 - 0.073) = 0.127 g`
44 g of `CO_(2) = 1` mole
0.127 g of `CO_(2) = 1/44 xx 0.127` moles
Moles of `CO_(2)` left `=0.127/44 = 2.89 xx 10^(-3)`
So, `2.89 xx 10^(-3)` moles of `CO_(2)` are left.