Question

# Determine the number of moles of $\mathrm{N}$ atoms in $43.5 \mathrm{~g}$ of $\mathrm{Mg}\left(\mathrm{NO}_3\right)_2$

Solutions

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Given: Mass of $\mathrm{Mg(NO_{3})_{2}}$ = 43.5 g

(1) First, calculate for the molar mass of $\mathrm{Mg(NO_{3})_{2}}$ as shown below:

$\text{Molar mass = Mg} + \mathrm{N_{2}} + \mathrm{O_{6}}$

$\text{Molar mass = 24.31 g + 2 (14.01 g) + 6 (16.00 g)}$

$\text{Molar mass = 148.3 g}$