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

Answer:

This means Ca(NO3)2 is the limiting reactant and Li3PO4 is the reactant in excess. LiNO3 and Ca3(PO4)2 are the products. Option C is correct.

Explanation:

Step 1: Data given

Number of moles Ca(NO3)2 = 3.4 moles

Number of moles Li3PO4 = 2.4 moles

Molar mass of Ca(NO3)2 = 164.09 g/mol

Molar mass of Li3PO4 =115.79 g/mol

Step 2: The balanced equation

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

Step 3: Calculate the limiting reactant

For 3 moles Ca(NO3)2 we need 2 moles Li3PO4 to produce 6 moles LiNO3 and 1 mol of Ca3(PO4)2

Ca(NO3)2 is the limiting reactant. It will completely be consumed. (3.4 moles). Li3PO4 is in excess. There will react 2/3 * 3.4 = 2.27 moles

There will remain 2.4 - 2.27 : 0.13 moles Li3PO4

This means Ca(NO3)2 is the limiting reactant and Li3PO4 is the reactant in excess. LiNO3 and Ca3(PO4)2 are the products. Option C is correct.