Limiting reactant problem?

Question

The following reaction is taking place: 13.6 grams of Lithium Carbonate reacts with 12.2 grams of sodium nitrate. Write out the balanced equation and show work to prove which reactant is the limiting reactant.

PLEASE SHOW WORK so I know how you did it!

Answer 1

Li2CO3 + 2 NaNO3 -> Na2CO3 + 2 LiNO3

Calculate molar mass of each and then find out the number of moles.

Answer 2

the equation above is correct.

Step 1: write reactants. Do it in pieces
Lithium carbonate, you must pair Li^+1 with CO3^-2 therefore you need 2 lithiums for every 1 carbonate
sodium nitrate: Na^+1 and NO3^-1 so you need a 1:1 ratio of those two
Step 2: write products: this is a double replacement reaction, so species do-see-do and switch partners. You never ever pair +/+ or -/- species together, so switch the anions around, since carbonate was paired with lithium as a reactant, on the product side it will be paired with sodium. Since nitrate was paired with sodium on the reactant side, it will then be paired with Li on the product side.
Approach writing the formulas the same way:
Na^+1 and CO3^-2 so you need 2 sodiums for every 1 carbonate
Li^+1 NO3^-1 so you need 1:1

Step 3: BALANCE the chemical reaction - if you do not do this you cannot answer anything about the reaction - in order to do calculations with chemical reactions, you need the correct stoichiometry

Li2CO3 + 2NaNO3 -------> 2LiNO3 + Na2CO3

Step 4: A limiting reactant problem: a sign that you have a limiting reactant problem is that you are given amounts of allllll the reactants. That means you MUST do a calculation with every amount of reactant that you have. The limiting reactants is NOT the species that is present in the smallest amount. It is the species that PRODUCES the smallest amount of product. After it has produced that amount of product, there isn't any more of that species left, so no more product can be made. Pick a product - any product. All calculations should be done based on forming that ONE AND ONLY one product. You pick ONE product so you can make a comparison of your reactants! PICK ONE.

I am going to pick LiNO3. No reason, it does NOT matter which product you pick, your limiting reagent will NOT change.

I need to know which species, the litium carbonate, or the sodium nitrate makes the least amount of product (lithium nitrate).

Therefore, I need to convert my grams of lithium carbonate into grams of lithium nitrate

AND I need to convert grams of sodium nitrate into grams of lithium nitrate

I need the following: in order to convert between grams and moles you need the molar mass of the species

In order to convert between chemical species in a chemical reaction I need the stoichiometry (the coefficients) in the balanced chemical reaction).

So:

13.6 g Li2CO3 x (1 moleLi2CO3/66.95 g Li2CO3) x (2 mole LiNO3/1 mole Li2CO3) x 68.95 g LiNO3/1molLiNO3)

Li2CO3 makes 28.0 grams of LiNO3

grams to moles rct -> mole to mole ratio -> moles to grams pdt

repeat the SAME process with the next reactant

12.2 g NaNO3 x (1 mol NaNO3/53.00 g NaNO3) x (2mol LiNO3/2 mole NaNO3) x (68.95 g LiNO3/1mole LiNO3)

NaNO3 makes 15.9 grams of LiNO3

again, grams rct to moles rct -> mole to mole ratio -> moles pdt to grams pdt

Since the NaNO3 makes the SMALLEST amount of product

NaNO3 is the limiting reagent

Answer 3

Find the number of moles of each of the two reactants, that's the amount of grams given/molecular weight. Then divide these moles by the coefficient in front of the respective species in the reaction above. The number that comes out smallest is the amount of limiting reactant used in the reaction.

Answer 4

Li2CO3 + 2 NaNO3 ---> 2 LiNO3 + Na2CO3

Li2CO3 = 73.8915 g/mol
NaNO3 = 84.9948 g/mol

(13.6 g Li2CO3) / (73.8915 g/mol Li2CO3)
= 0.184 mol Li2CO3

(12.2 g NaNO3) / (84.9948 g/mol NaNO3)
= 0.144 mol NaNO3

From the balanced equation... You need 2 moles of sodium nitrate for every 1 mole of lithium carbonate. So, to react all 0.184 moles of lithium carbonate, you'll need 2 * 0.184 = 0.368 moles of sodium nitrate. However, you only have 0.144 moles of sodium nitrate.

Thus, sodium nitrate (NaNO3) is your limiting reactant.