Question

Silver sulfate is precipitated from solution according to the following reaction 2 AgNO3 (aq) Na2SO4 (aq) --> Ag2SO4 (s) 2 NaNO3 (aq) If 3.80 moles of AgNO3 and 3.05 moles of Na2SO4 are present initially, how many moles of excess reactant will remain after precipitation is complete

Answer 1

Answer:

1.15 moles of excess reactant will remain after precipitation is complete.

Explanation:

The balanced reaction is:

2 AgNO₃ (aq) + Na₂SO₄ (aq) → Ag₂SO₄ (s) + 2 NaNO₃ (aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

• AgNO₃: 2 moles
• Na₂SO₄: 1 mole
• Ag₂SO₄: 1 mole
• NaNO₃: 2 moles

Then you can apply the following rule of three:: if by stoichiometry 2 moles of AgNO₃ reacts with 1 mole of Na₂SO₄, 3.80 moles of AgNO₃ reacts with how much moles of Na₂SO₄?

$amount of moles of Na_{2}SO_{4} =\frac{1mole of Na_{2}SO_{4} * 3.80 moles of AgNO_{3} }{2 mols of AgNO_{3} }$

amount of moles of Na₂SO₄= 1.9 moles

But 3.05 moles of Na₂SO₄ are available. Since you have more moles than you need to react with 3.80 moles of AgNO₃, Na₂SO₄ will be the excess reagent.

To calculate the amount of excess reagent that will remain, you must make the difference between the amount you initially have and the amount that reacts:

3.05 moles - 1.9 moles= 1.15 moles

1.15 moles of excess reactant will remain after precipitation is complete.