Question

How many grams of Al3+ are present in 1.98 grams of aluminum cyanide?

Answer 1

0.509 grams Al³⁺

Step-by-step 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:

`\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}*27.0grams\text{ }Al^(3+) \\ \\ Grams\text{ }of\text{ }Al^(3+)=0.509\text{ }grams\text{ }Al^(3+) \end{gathered}`

Thus, 0.509 grams of Al³⁺ are present in 1.98 grams of aluminum cyanide.