Answer
0.509 grams Al³⁺
Explanation
Given:
Mass of aluminum cyanide = 1.98 grams
Required: To find the grams of Al³⁺ present in 1.98 grams of aluminum cyanide.
Step-by-step solution:
The grams of 1 mole of aluminum cyanide (Al(CN)₃) = 105.03 grams
This implies that 27.0 g Al³⁺ is present in 105.0 grams of Al(CN)₃
So, the grams of Al³⁺ present in 1.98 grams of Al(CN)₃ will be:
[tex]\begin{gathered} Grams\text{ }of\text{ }Al^{3+}=\frac{1.98\text{ }grams\text{ }Al(CN)_3}{105.0\text{ }grams\text{ }Al(CN)_3}\times27.0grams\text{ }Al^{3+} \\ \\ Grams\text{ }of\text{ }Al^{3+}=0.509\text{ }grams\text{ }Al^{3+} \end{gathered}[/tex]
Thus, 0.509 grams of Al³⁺ are present in 1.98 grams of aluminum cyanide.
