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%

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

the answer is minerals.