Question

1.

a) If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced? (0.74577)

b) If 0.726 g of sulfuric acid is produced in the laboratory experiment, what is the percent yield of sulfuric acid? (97.3190)

2.

If 3.2 moles of sodium sulfate is used in the reaction, how many formula units of phosphoric acid is needed to completely react?
(1.28427×1024

Answer 1

1. a) 0.745 g of H₂SO₄

b) 97.4%

2. 1.28×10²⁴ formula units

Step-by-step explanation:

First of all, we determine the equations:

3Na₂SO₄ + 2H₃PO₄ → 3H₂SO₄ + 2Na₃PO₄

We determine, the moles of sodium sulfate. (As the acid is in excess, we work with the salt)

1.08 g . 1mol/ 142.04g = 0.00760 moles.

Ratio is 3:3. So 3 moles of salt can produce 3 moles of sulfuric acid,

Then, 0.00760 moles will produce the same amount of acid.

We convert the moles to mass → 0.00760 mol . 98 g/mol = 0.745 g

Percent yield = (Produced yield /Theoretical yield) . 100

(0.726 g / 0.745 g) . 100 = 97.4 %

2. 3 moles of sodium sulfate need 3 moles of phosphoric acid to react

Then, 3.2 moles of salt will react to (3.2 . 2)/3 = 2.13 moles

Let's count the formula units:

2.13 mol . 6.02×10²³ formula units /mol = 1.28×10²⁴ formula units.