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# Calculate each of the following quantities:(a) Mass (g) of $6.44 \times 10^{-2} \mathrm{mol}$ of $\mathrm{MnSO}_{4}$(b) Amount (mol) of compound in 15.8 $\mathrm{kg}$ of $\mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{3}$(c) Number of $\mathrm{N}$ atoms in 92.6 $\mathrm{mg}$ of $\mathrm{NH}_{4} \mathrm{NO}_{2}$

## a) $\mathrm{m}=9,725 \mathrm{g}$b) $\mathrm{n}\left(\mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{3}\right)=44,61 \mathrm{mol}$c) $\mathrm{N}(\mathrm{N})=1,74^{*} 10^{21}$ atoms

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Chemical reactions and Stoichiometry

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