![\rm K_3[Fe(CN)_6]](/tpl/images/0722/9282/07693.png)
(potassium ferricyanide) is approximately 
.![\begin{aligned} & M(\mathrm{K_3[Fe(CN)_6]}) \\ &\approx 3 \times 39 + 56 + 6 \times (12 + 14) \\ &= \rm 329 \; g \cdot mol^{-1}\end{aligned}](/tpl/images/0722/9282/fc522.png)
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atom in every one mole of 
(again, from the relative atomic mass data of the question.) ![\begin{aligned}& \text{Mass percentage of $\mathrm{Fe}$ in $\mathrm{K_3[Fe(CN)_6]}$} \\ &= \frac{\text{Mass of $\mathrm{Fe}$ in one mole of $\mathrm{K_3[Fe(CN)_6]}$ formula units}}{\text{Mass of one mole of $\mathrm{K_3[Fe(CN)_6]}$ formula units}}\times 100\% \\ &= \frac{M(\mathrm{Fe}) \times (\text{Number of $\mathrm{Fe}$ atoms in each $\mathrm{K_3[Fe(CN)_6]}$ formula unit)}}{M(\mathrm{K_3[Fe(CN)_6]})} \times 100\%\end{aligned}](/tpl/images/0722/9282/ba0e8.png)
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Assignment
1. Calculate the percentage of iron in K3 Fe (CN)
Chemistry, 16.08.2020 01:01 letsgivepece2309
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Date
Page
Assignment
1. Calculate the percentage of iron in K3 Fe (CN)
[K= 39, Fe =56, C= 12, N =14]
Answers: 1
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1. Calculate the percentage of iron in K3 Fe (CN)
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