Question

What is the limiting reactant for the following reaction given we have 3.4 moles of Ca(NO3)2 and 2.4 moles of Li3PO4?Reaction: 3Ca(NO3)2 + 2Li3PO4 ---> 6LiNO3 + Ca3(PO4)2Answer (show work please?)a.Ca3(PO4)2b.Li3PO4c.Ca(NO3)2d.LiNO3

Answer 1

Answer:

The answer to your question is   letter c. Ca(NO₃)₂

Explanation:

Data

Ca(NO₃)₂  = 3.4 moles

Li₃PO₄ = 2.4 moles

Reaction

                   3 Ca(NO₃)₂  +  2 Li₃PO₄   ⇒   6 LiNO₃  +  Ca₃(PO₄)₂

Process

Calculate the proportion theoretical and experimental of reactants and compare these proportions.

Theoretical proportion

                   Ca(NO₃)₂ / Li₃PO₄  = $\frac{3 moles}{2 moles}$ = 1.5

Experimental proportion

                   Ca(NO₃)₂ / Li₃PO₄  = $\frac{3.4}{2.4} = 1.42$

As the experimental proportion is lower than the theoretical proportion we conclude that the amount of Li₃PO₄ increased in the experiment so the limiting reactant is Ca(NO₃)₂.

Answer 2

The limiting reactant would be Ca(NO3)2

From the equation of the reaction:

3Ca(NO3)2 + 2Li3PO4 ---> 6LiNO3 + Ca3(PO4)2

The mole ratio of Ca(NO3)2 to Li3PO4 is 3:2. That is, for every 3 moles of Ca(NO3)2, 2 moles of Li3PO4 is required.

If there are 3.4 moles of Ca(NO3)2, the moles of Li3PO4 required would be:

       3.4 x 2/3 = 2.27 moles

However, 2.4 moles of Li3PO4 is available. This means that  Li3PO4 is in excess and Ca(NO3)2 is limiting.

More on limiting reagents can be found here: https://brainly.com/question/11848702