I think there is a typo because I've never seen HSO4 2- before in my life. It should be HSO4-. For that, H is 1+ and each Oxygen is 2-0 totaling 8-. So the oxidation state of sulfur +1 - 8 = 7

So the oxidation state of sulfur is +6

7 0

3 years ago

Which of the following is a product in the chemical equation N + O2 = NO2

zlopas [31]

The answer is 1. NO2

3 0

3 years ago

Which of the following correctly represents the radioactive decay of cesium-137?<br><br>

pishuonlain [190]

What are the answers ??

5 0

3 years ago

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

vfiekz [6]

Answer: 0.745 g of $H_2SO_4$ will be produced from 1.08 g of sodium sulfate

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of} Na_2SO_4=\frac{1.08g}{142.04g/mol}=0.0076moles$

$3Na_2SO_4+2H_3PO_4\rightarrow 2Na_3PO_4+3H_2SO_4$

$Na_2SO_4$ is the limiting reagent as it limits the formation of product and $H_3PO_4$ is the excess reagent.

According to stoichiometry :

3 moles of $Na_2SO_4$ produce = 3 moles of $H_2SO_4$

Thus 0.0076 moles of $Na_2SO_4$ will require= $\frac{3}{3}\times 0.0076=0.0076moles$ of $H_2SO_4$

Mass of $H_2SO_4=moles\times {\text {Molar mass}}=0.0076moles\times 98.1g/mol=0.745g$

Thus 0.745 g of $H_2SO_4$ will be produced from 1.08 g of sodium sulfate

3 0

2 years ago