2FePO4+3 Na2so4--->Fe2(so4) + 2 Na3Po4 if i ise 25grams of iron iii phosphate 18.5 g iton what is my percent yeild?

56%

Explanation:

If I use 25 grams of iron (III) phosphate and obtain 18.5 g of iron (III) sulfate, what is my percent yield?

Step 1: Write the balanced equation

2 FePO₄ + 3 Na₂SO₄ ⇒ Fe₂(SO₄)₃ + 2 Na₃PO₄

Step 2: Calculate the theoretical yield of Fe₂(SO₄)₃

The mass ratio of FePO₄ to Fe₂(SO₄)₃ is 301.64:399.88.

25 g FePO₄ × 399.88 g Fe₂(SO₄)₃/301.64 g FePO₄ = 33 g Fe₂(SO₄)₃

Step 3: Calculate the percent yield of Fe₂(SO₄)₃

We will use the following expression.

%yield = (experimental yield/theoretical yield) × 100%

%yield = (18.5 g/33 g) × 100% = 56%

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."

it forms an anion that has a larger radius. anions have a negative charge and are larger than the original atom.