What is the limiting reactant for the following reaction if there are 3.4 moles of Ca(NO2)2 and 2.4 moles of Li3PO4?
3Ca(NO2)2 + 2Li3PO4 --> 6LiNO2 + Ca3(PO4)2
A)Ca(NO2)2
B) Li3PO4
C) LiNO2
D) Ca3(PO4)2
to find LR divide the amount by the coefficient in the equation, whichever is lower, is the Limiting Reactant (LR)
3.4/3 = 1.13
2.4/2 = 1.2
So the Ca(NO2)@ is limiting (A)
Ok Good thank you, I just get confused as to the denominator in the LR's...Can you use any of the coefficients in the reactants?
To determine the limiting reactant in a chemical reaction, you need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
The balanced equation for the reaction is:
3Ca(NO2)2 + 2Li3PO4 → 6LiNO2 + Ca3(PO4)2
First, let's calculate the number of moles of 3Ca(NO2)2 and 2Li3PO4:
Number of moles of Ca(NO2)2 = 3.4 moles
Number of moles of Li3PO4 = 2.4 moles
Next, we need to determine the stoichiometric ratio for both reactants in the balanced equation. The stoichiometric ratio is the ratio of moles between the reactants in the balanced equation. In this case, it is 3:2 for Ca(NO2)2 and Li3PO4.
Now, we can compare the actual number of moles to the stoichiometric ratio for each reactant:
Ca(NO2)2:
Actual ratio = 3.4 moles / 3 = 1.13
Li3PO4:
Actual ratio = 2.4 moles / 2 = 1.2
To find the limiting reactant, we need to compare the actual ratios. The limiting reactant is the reactant with the smallest actual ratio. In this case, the actual ratio for Ca(NO2)2 is 1.13, and the actual ratio for Li3PO4 is 1.2.
Since the actual ratio for Ca(NO2)2 is smaller, it will be completely consumed first, making it the limiting reactant. Therefore, the answer is:
A) Ca(NO2)2