Which of the following contains the maximum n | vedclass

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(A) To find the number of atoms,we first calculate the number of moles and then multiply by the number of atoms per molecule and Avogadro's number $(N

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$4 \ g$ of $H_2$

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(A) To find the number of atoms,we first calculate the number of moles and then multiply by the number of atoms per molecule and Avogadro's number $(N_A)$.$1$. For $4 \ g$ of $H_2$: Moles = $4 \ g / 2 \ g/mol = 2 \ mol$. Atoms = $2 \ mol 	imes 2 \ (atoms/molecule) 	imes N_A = 4N_A$.$2$. For $16 \ g$ of $O_2$: Moles = $16 \ g / 32 \ g/mol = 0.5 \ mol$. Atoms = $0.5 \ mol 	imes 2 \ (atoms/molecule) 	imes N_A = 1N_A$.$3$. For $28 \ g$ of $N_2$: Moles = $28 \ g / 28 \ g/mol = 1 \ mol$. Atoms = $1 \ mol 	imes 2 \ (atoms/molecule) 	imes N_A = 2N_A$.$4$. For $18 \ g$ of $H_2O$: Moles = $18 \ g / 18 \ g/mol = 1 \ mol$. Atoms = $1 \ mol 	imes 3 \ (atoms/molecule) 	imes N_A = 3N_A$.Comparing the values,$4 \ g$ of $H_2$ contains the maximum number of atoms $(4N_A)$.

Which of the following contains the maximum number of atoms?

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