Question

The following data was collected when a reaction was performed experimentally in the laboratory. What is the maximum amount of NaNO3?And how did you find the amount?

Answer 1

Step-by-step 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:

`Al\left(NO_3\right)_3+3NaCl\rightarrow3NaNO_3+AlCl_3`

Number of moles of NaNO3 from Al(NO3)3:

`\begin{gathered} n(NaNO_3)\text{ = n\lparen Al\lparen NO}_3)_3)*\frac{n(NaNO_3)}{n(\text{Al\lparen NO}_3)_3)} \\ \\ \text{ = 4}*(3)/(1) \\ \\ \text{ = 12 mol} \end{gathered}`

Number of moles of NaNO3 from NaCl:

`\begin{gathered} n(NaNO_3)=\text{ n\lparen NaCl\rparen}*(n(NaNO_3))/(n(NaCl)) \\ \\ \text{ = 9}*(3)/(3) \\ \\ \text{ = 9 mol} \end{gathered}`

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.