Question #10dc1

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Question #10dc1

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Answer 1 · 2017-04-10T16:39:44

$= 0.340$ $=0.340$ $m o l$ $mol$

Explanation:

Using the periodic table, the molar mass of $F e {(C_{2} H_{3} O_{2})}_{3}$ $Fe(C_2H_3O_2)_3$ is approximatly $232.98 \frac{g}{m o l}$ $232.98g/(mol)$

Using the molar mass and the grams given, you can solve for moles,

$79.3 g \cdot \frac{1}{232.98} \frac{m o l}{g}$ $79.3g*1/232.98(mol)/g$

$79.3cancelg*1/232.98(mol)/cancelg$

$79.3/232.98mol$

$=0.340$ $mol$

Or you could of used the equation,

$n=m/(mm)$

Where $n$ is moles, $m$ is mass in $grams$ and $mm$ is molar mass in $(grams)/(mol)$

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Question #10dc1

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