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Nitella [24]
3 years ago
10

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

Chemistry
1 answer:
vfiekz [6]3 years ago
3 0

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

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