From 392 , mg of H_2SO_4,1.204 times 10^{21} | vedclass

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(A) The molar mass of $H_2SO_4$ is $98 \, g/mol$.Total moles of $H_2SO_4$ initially $= \frac{392 	imes 10^{-3} \, g}{98 \, g/mol} = 4 	imes 10^{-3}

Vedclass · Accepted Answer

$2.0 	imes 10^{-3}$

Vedclass · Answer

(A) The molar mass of $H_2SO_4$ is $98 \, g/mol$.Total moles of $H_2SO_4$ initially $= \frac{392 	imes 10^{-3} \, g}{98 \, g/mol} = 4 	imes 10^{-3} \, mol$.Moles of $H_2SO_4$ removed $= \frac{1.204 	imes 10^{21}}{6.02 	imes 10^{23} \, mol^{-1}} = 0.2 	imes 10^{-2} \, mol = 2 	imes 10^{-3} \, mol$.Moles of $H_2SO_4$ left $= (4 	imes 10^{-3}) - (2 	imes 10^{-3}) = 2.0 	imes 10^{-3} \, mol$.

From $392 \, mg$ of $H_2SO_4$,$1.204 \times 10^{21}$ molecules are removed. How many moles of $H_2SO_4$ are left?

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