Answer:
3 mol AlCl₃.
Explanation:
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In this case, according to the specified reactants and products, it is possible to set up the following balanced chemical reaction:
[tex]Al(NO_3)_3+3NaCl\rightarrow 3NaNO_3+AlCl_3[/tex]
Whereas we evidence the 1:3 mole ratio between aluminum nitrate and sodium chloride; thus, since different moles were reacting, we need to identify the limiting reactant by computing the moles of AlCl₃ produced by each reactant as follows:
[tex]n_{AlCl_3}^{by\ Al(NO_3)_3}=4molAl(NO_3)_3*\frac{1molAlCl_3}{1molAl(NO_3)_3} =4molAlCl_3\\\\n_{AlCl_3}^{by\ NaCl}=9molNaCl*\frac{1molAlCl_3}{3molNaCl} =3molAlCl_3[/tex]
Thus, we infer that NaCl is the limiting reactant as it produces the fewest moles of AlCl₃; consequently the produced amount of this product is 3 mol.
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