A l ( N O 3 ) 3 ( s ) ⟶ A l 3 + ( a q ) + 3 N O 3 − ( a q ) Calculate the moles of nitrate ions in 20.77mL of a 0.71M aluminum nitrate solution. Hint: multiply by the mole ratio

Al(NO₃)₃ (s)  ⟶ Al³⁺ (aq) + 3NO³⁻ (aq)

To calculate the number of moles of nitrate ions in 20.77 ml of a 0.71 M aluminum nitrate.

1 ml equals to 10^-3 L therefore 20.77 ml equals to 20.77×10^-3 L

Calculate the number of moles of aluminum nitrate in 20.77×10^-3 L solution

$\begin{array}{l}\text{Number of moles}=\left(\text{Concentration}\right)\left(\text{Volume of solution}\right)\\ \text{Number of moles}=\left(0.7\text{ M}\right)\left(20.77\times {10}^{-3}\text{ L}\right)\\ \text{Number of moles}=14.539\times {10}^{-3}\text{ mol}\end{array}$

Thus, number of moles of aluminum nitrate in the solution is 14.539×10^-3 mol