Question

How many grams of K3PO4 will be formed from 134.5 moles of Na3PO4?

Answer 1

To find the amount of K3PO4 formed, we use the stoichiometry of the balanced equation to set up a proportion and solve for x. The amount of K3PO4 formed from 134.5 moles of Na3PO4 is 67.25 moles or 14,263.98 grams.

Step-by-step explanation:

To find the amount of K3PO4 formed from 134.5 moles of Na3PO4, we need to use the stoichiometry of the balanced equation.

The balanced equation is:

2Na3PO4 + 3K2SO4 → 6Na2SO4 + K3PO4

This equation shows that for every 2 moles of Na3PO4, we will get 1 mole of K3PO4. Therefore, we can set up a proportion:

{{134.5 moles Na3PO4}/{2 moles Na3PO4}} = {{x moles K3PO4}/ {1 mole K3PO4}}

Solving for x gives us x = 67.25 moles K3PO4.

Finally, since the molar mass of K3PO4 is 212.27 g/mol, we can convert moles to grams:

{{67.25 moles K3PO4} × {212.27 g/mol K3PO4}} = 14,263.98 g K3PO4

Answer 2

The balanced chemical equation for the reaction between Na3PO4 and KCl is:

2 Na3PO4 + 3 KCl → K3PO4 + 3 NaCl

According to the equation, 2 moles of Na3PO4 react with 3 moles of KCl to form 1 mole of K3PO4. Therefore, the number of moles of K3PO4 formed from 134.5 moles of Na3PO4 can be calculated as:

134.5 moles Na3PO4 x (1 mole K3PO4 / 2 moles Na3PO4) = 67.25 moles K3PO4

To convert moles of K3PO4 to grams, we need to use its molar mass. The molar mass of K3PO4 can be calculated as:

3(39.10 g/mol K) + 1(30.97 g/mol P) + 4(16.00 g/mol O) = 212.27 g/mol

Therefore, the mass of K3PO4 formed can be calculated as:

67.25 moles K3PO4 x (212.27 g/mol K3PO4) = 14,259.98 g

Rounding off to 2 decimal places, the answer is 14,260.00 grams of K3PO4.