Explanation:
We are given: moles of Al(NO3)3 = 4 mol
: moles of NaCl = 9 mol
We know: molar mass of Al(NO3)3 = 212.996 g/mol
: molar mass of NaCl = 58.44 g/mol
: molar mass of NaNO3 = 84.9947 g/mol
The balanced chemical equation is given as:
[tex]Al\left(NO_3\right)_3+3NaCl\rightarrow3NaNO_3+AlCl_3[/tex]
Number of moles of NaNO3 from Al(NO3)3:
[tex]\begin{gathered} n(NaNO_3)\text{ = n\lparen Al\lparen NO}_3)_3)\times\frac{n(NaNO_3)}{n(\text{Al\lparen NO}_3)_3)} \\ \\ \text{ = 4}\times\frac{3}{1} \\ \\ \text{ = 12 mol} \end{gathered}[/tex]
Number of moles of NaNO3 from NaCl:
[tex]\begin{gathered} n(NaNO_3)=\text{ n\lparen NaCl\rparen}\times\frac{n(NaNO_3)}{n(NaCl)} \\ \\ \text{ = 9}\times\frac{3}{3} \\ \\ \text{ = 9 mol} \end{gathered}[/tex]
Therefore, NaCl is a limiting reagent.
Answer:
The maximum amount of NaNO3 is 9 mol.
By balancing the chemical equation of the given reactants and products. And then use molar ratios and the number of moles to find the limiting reagent.
