For the reaction
3 FeCl2 + 2 Na3PO4 → Fe3(PO4)2 + 6 NaCl ,
what is the maximum amount of Fe3(PO4)2
which could be formed from 11.26 g of FeCl2
and 13.34 g of Na3PO4?
Oh, buckle up! We're going on a fun-filled chemical adventure! To find the maximum amount of Fe3(PO4)2 formed, we need to determine the limiting reactant.
Let's start with FeCl2. The molar mass of FeCl2 is 126.75 g/mol. So, 11.26 g of FeCl2 is equal to 11.26 g / 126.75 g/mol, which is about 0.089 mol.
Now, let's move on to Na3PO4. The molar mass of Na3PO4 is 163.94 g/mol. Therefore, 13.34 g of Na3PO4 is equal to 13.34 g / 163.94 g/mol, which is roughly 0.081 mol.
Oh no, it seems like we're dealing with a runaway circus of moles here! Since we need 2 moles of Na3PO4 for every 3 moles of FeCl2, let's do some division.
Dividing the moles of FeCl2 (0.089) by the stoichiometric coefficient (3) gives us 0.030 mol. Now, let's divide the moles of Na3PO4 (0.081) by the stoichiometric coefficient (2): we get 0.0405 mol.
As the audience is holding their breath, we see that 0.030 mol of FeCl2 can react with 0.015 mol of Na3PO4 to form Fe3(PO4)2. But oh dear, we only have 0.0105 mol of Na3PO4 remaining unused!
So, the limiting reactant is Na3PO4, bringing the maximum amount of Fe3(PO4)2 formed to 0.015 mol. Now, to find the mass, we multiply the moles (0.015 mol) by the molar mass of Fe3(PO4)2 (357.53 g/mol), which gives us a dazzling result of 5.36 g.
Voila! The maximum amount of Fe3(PO4)2 that could be formed is approximately 5.36 g. And the audience goes wild with applause!
To determine the maximum amount of Fe3(PO4)2 that can be formed, we will use the concept of limiting reactants. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Step 1: Convert mass to moles
Using the molar masses, we can convert the given masses of FeCl2 and Na3PO4 to moles.
Molar mass of FeCl2 = (55.85 g/mol) + (2 * 35.45 g/mol) = 126.75 g/mol
Molar mass of Na3PO4 = (22.99 g/mol) + (3 * 16.00 g/mol) + (4 * 15.99 g/mol) = 163.94 g/mol
Moles of FeCl2 = 11.26 g / 126.75 g/mol ≈ 0.089 mol
Moles of Na3PO4 = 13.34 g / 163.94 g/mol ≈ 0.081 mol
Step 2: Determine the limiting reactant
To determine the limiting reactant, we compare the stoichiometric ratios of the reactants.
From the balanced equation, the ratio of FeCl2 to Fe3(PO4)2 is 3:1
From the balanced equation, the ratio of Na3PO4 to Fe3(PO4)2 is 2:1
The smaller ratio is 0.081 mol of Na3PO4 to Fe3(PO4)2.
Therefore, Na3PO4 is the limiting reactant.
Step 3: Calculate the maximum amount of Fe3(PO4)2 formed
From the balanced equation, we know that the stoichiometric ratio of Na3PO4 to Fe3(PO4)2 is 2:1.
Moles of Fe3(PO4)2 = 0.081 mol (Na3PO4) * (1 mol Fe3(PO4)2 / 2 mol Na3PO4) = 0.0405 mol
Step 4: Convert moles to grams
Using the molar mass of Fe3(PO4)2:
Molar mass of Fe3(PO4)2 = (55.85 g/mol) + (2 * 31.00 g/mol) + (8 * 16.00 g/mol) = 357.74 g/mol
Mass of Fe3(PO4)2 = 0.0405 mol * 357.74 g/mol ≈ 14.5 g
Therefore, the maximum amount of Fe3(PO4)2 that can be formed from 11.26 g of FeCl2 and 13.34 g of Na3PO4 is approximately 14.5 grams.
To calculate the maximum amount of Fe3(PO4)2 that could be formed from given amounts of FeCl2 and Na3PO4, we need to use stoichiometry and the concept of limiting reactants.
1. Find the molar mass of FeCl2 and Na3PO4:
- Molar mass of FeCl2 = (1 × atomic mass of Fe) + (2 × atomic mass of Cl)
- Molar mass of Na3PO4 = (3 × atomic mass of Na) + atomic mass of P + (4 × atomic mass of O)
2. Convert the given masses of FeCl2 and Na3PO4 to moles:
- Moles of FeCl2 = given mass of FeCl2 / molar mass of FeCl2
- Moles of Na3PO4 = given mass of Na3PO4 / molar mass of Na3PO4
3. Write a balanced equation for the reaction and determine the stoichiometric ratio:
- From the balanced equation, we can see that the ratio of FeCl2 to Fe3(PO4)2 is 3:1. This means that for every 3 moles of FeCl2, we get 1 mole of Fe3(PO4)2.
4. Identify the limiting reactant:
- To determine the limiting reactant, compare the moles of FeCl2 and Na3PO4 obtained in step 2.
- Whichever reactant has a smaller number of moles is the limiting reactant.
5. Calculate the maximum moles (or mass) of Fe3(PO4)2 formed:
- If FeCl2 is the limiting reactant, use the mole ratio from the balanced equation to calculate moles (or mass) of Fe3(PO4)2 formed.
- If Na3PO4 is the limiting reactant, use the mole ratio and the number of moles (or mass) of Na3PO4 to calculate moles (or mass) of Fe3(PO4)2 formed.
6. Finally, convert moles (or mass) of Fe3(PO4)2 to grams, if necessary.
By following these steps, you can calculate the maximum amount of Fe3(PO4)2 that can be formed from the given masses of FeCl2 and Na3PO4.
This is a limiting reagent problem. You know that because amounts are given for BOTH reactants.
Convert 11.26 g FeCl2 to mols. mols = g/molar mass. Approximately 0.09
Convert 13.34g Na3PO4 to mols.
Approximately 0.08
Convert mols FeCl2 to mols of the product using the coefficients in the balanced equation. About 0.09 x 1/3 = about 0.03
Convert mols Na3PO4 to mols of the product. Same procedure. About 0.08 x 1/2 = 0.04.
You will note that you have two values for the product; of course, both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value. Make sure to confirm all of the values above because I've estimated them.
Then mols product x molar mass product = grams product.