Answer:
About 2.20 moles.
Explanation:
We want to determine the amount of moles of aluminum cyanide in 231 grams of the substance.
To do so, we can convert from moles to grams using its molar mass.
Find the molar mass of Al(CN)₃:
[tex]\displaystyle \begin{aligned} \mathcal{M}_{\text{Al(CN)$_3$} } & = \left( \underbrace{26.98}_{\text{Al}} + 3(\underbrace{12.01}_{\text{C}}) + 3(\underbrace{14.01}_{\text{N}})\right) \text{ g/mol} \\ \\ & =105.04\text{ g/mol} \end{aligned}[/tex]
With the initial value, multiply:
[tex]\displaystyle 231 \text{ g Al(CN)$_3$} \cdot \frac{1 \text{ mol Al(CN)$_3$}}{105.04\text{ g Al(CN)$_3$}} = 2.20\text{ mol Al(CN)$_3$}[/tex]
In conclusion, there are about 2.20 moles of aluminum cyanide in 231 grams of the substance.
