If 10^{21} molecules are removed from 100 , m | vedclass

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(C) Initial mass of $CO_2 = 100 \, mg = 0.1 \, g$. Molar mass of $CO_2 = 12 + (2 	imes 16) = 44 \, g/mol$. Initial moles of $CO_2 = \frac{0.1 \, g}{4

Vedclass · Accepted Answer

$6 	imes 10^{-4}$

Vedclass · Answer

(C) Initial mass of $CO_2 = 100 \, mg = 0.1 \, g$. Molar mass of $CO_2 = 12 + (2 	imes 16) = 44 \, g/mol$. Initial moles of $CO_2 = \frac{0.1 \, g}{44 \, g/mol} \approx 2.27 	imes 10^{-3} \, mol$. Moles of $CO_2$ removed = $\frac{10^{21}}{6.022 	imes 10^{23}} \approx 1.66 	imes 10^{-3} \, mol$. Remaining moles of $CO_2 = (2.27 	imes 10^{-3}) - (1.66 	imes 10^{-3}) = 0.61 	imes 10^{-3} \, mol = 6.1 	imes 10^{-4} \, mol$. Rounding to the nearest provided option,the answer is $6 	imes 10^{-4} \, mol$.

If $10^{21}$ molecules are removed from $100 \, mg$ of $CO_2$,how many moles of $CO_2$ remain?

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