For the reaction
? FeCl2 + ? Na3PO4 → ? Fe3(PO4)2 + ? NaCl
what is the maximum mass of Fe3(PO4)2 that
could be formed from 5.25 g of FeCl2 and
1.62 g of Na3PO4?
Answer in units of g
Na cl
To find the maximum mass of Fe3(PO4)2 that can be formed, we need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed first and determines the maximum amount of product that can be formed.
Let's start by calculating the number of moles for each reactant using their molar masses:
Molar mass of FeCl2 (iron(II) chloride) = 126.75 g/mol
Molar mass of Na3PO4 (sodium phosphate) = 164.04 g/mol
Number of moles of FeCl2 = mass of FeCl2 / molar mass of FeCl2
Number of moles of FeCl2 = 5.25 g / 126.75 g/mol = 0.0414 mol
Number of moles of Na3PO4 = mass of Na3PO4 / molar mass of Na3PO4
Number of moles of Na3PO4 = 1.62 g / 164.04 g/mol = 0.00987 mol
Now, we can use the balanced chemical equation to determine the stoichiometry between FeCl2 and Fe3(PO4)2:
From the balanced equation:
1 FeCl2 : 1 Fe3(PO4)2
Therefore, the number of moles of Fe3(PO4)2 formed is equal to the number of moles of FeCl2.
Number of moles of Fe3(PO4)2 = 0.0414 mol
Finally, we can calculate the maximum mass of Fe3(PO4)2 that can be formed using its molar mass:
Molar mass of Fe3(PO4)2 (iron(III) phosphate) = 357.52 g/mol
Maximum mass of Fe3(PO4)2 = number of moles of Fe3(PO4)2 * molar mass of Fe3(PO4)2
Maximum mass of Fe3(PO4)2 = 0.0414 mol * 357.52 g/mol = 14.69 g
Therefore, the maximum mass of Fe3(PO4)2 that can be formed from 5.25 g of FeCl2 and 1.62 g of Na3PO4 is 14.69 g.
To determine the maximum mass of Fe3(PO4)2 that could be formed, we need to calculate the limiting reactant first. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's start by calculating the number of moles for FeCl2 and Na3PO4 using their respective molar masses.
The molar mass of FeCl2 (Iron(II) chloride) is:
Fe: 55.85 g/mol
Cl: 35.45 g/mol (x 2 since there are 2 chloride ions)
So, the molar mass of FeCl2 is 55.85 g/mol + 35.45 g/mol x 2 = 126.75 g/mol.
To find the number of moles, we divide the mass of FeCl2 by its molar mass:
Number of moles of FeCl2 = 5.25 g / 126.75 g/mol
Next, we calculate the number of moles for Na3PO4 (Sodium phosphate).
The molar mass of Na3PO4 is:
Na: 22.99 g/mol (x 3 since there are 3 sodium ions)
P: 30.97 g/mol
O: 16.00 g/mol (x 4 since there are 4 oxygen atoms)
So, the molar mass of Na3PO4 is 22.99 g/mol x 3 + 30.97 g/mol + 16.00 g/mol x 4 = 163.94 g/mol.
To find the number of moles, we divide the mass of Na3PO4 by its molar mass:
Number of moles of Na3PO4 = 1.62 g / 163.94 g/mol
Now we determine the limiting reactant.
For the balanced equation:
FeCl2 + Na3PO4 → Fe3(PO4)2 + 3NaCl
From the equation, we can see that the mole ratio between FeCl2 and Fe3(PO4)2 is 1:1.
So, the number of moles of Fe3(PO4)2 that can be formed is equal to the number of moles of FeCl2.
Now we need to calculate the maximum mass of Fe3(PO4)2 that can be formed.
The molar mass of Fe3(PO4)2 (Iron(III) phosphate) is:
Fe: 55.85 g/mol (x 3 since there are 3 iron atoms)
P: 30.97 g/mol (x 2 since there are 2 phosphate ions)
O: 16.00 g/mol (x 8 since there are 8 oxygen atoms)
So, the molar mass of Fe3(PO4)2 is 55.85 g/mol x 3 + 30.97 g/mol x 2 + 16.00 g/mol x 8 = 406.92 g/mol.
To find the maximum mass of Fe3(PO4)2, we multiply the number of moles of Fe3(PO4)2 by its molar mass:
Maximum mass of Fe3(PO4)2 = Number of moles of Fe3(PO4)2 x Molar mass of Fe3(PO4)2
Therefore, the maximum mass of Fe3(PO4)2 that could be formed from the given masses of FeCl2 and Na3PO4 is the same as the mass of FeCl2, which is 5.25 g. So, the answer is 5.25 g.
This is a limiting reagent problem; I know that because amounts are given for BOTH reactants.
6FeCl2 + 4Na3PO4 → 2Fe3(PO4)2 + 12NaCl
mols FeCl2 = grams/molar mass.
mols Na3PO4 = grams/molar mass
Convert mols FeCl2 to mols Fe3(PO4)2 using the coefficients in the balanced equation.
Convert mols Na3PO4 to mols Fe3(PO4)2.
It is likely these two values will be different which means one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for producing that number is the limiting reagent.
Use the smaller value and convert to grams. g = mols x molar mass.