Final answer:
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