Question

What volume, in mL, of a 4.69% AgNO3 solution is required to react exactly with 9.9 mL of 14.01% Na2SO4 solution? Hint: You will want to write a balanced reaction. The answer is 70.7345 but I'm not sure how to get to that

Answer 1

70.7345 mL. of a 4.69% AgNO3 solution is required to react exactly with 9.9 mL of 14.01% Na2SO4 solution.

Step-by-step explanation:

To calculate the volume of a 4.69% AgNO3 solution needed to react with 9.9 mL of 14.01% Na2SO4 solution, you need to write a balanced chemical equation and perform stoichiometric calculations. We are dealing with a precipitation reaction where AgNO3 reacts with Na2SO4 to form Ag2SO4 and NaNO3:

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

The reaction ratio for AgNO3 to Na2SO4 is 2:1. To find the moles of Na2SO4, you use the concentration and the volume of Na2SO4 solution:

Moles of Na2SO4 = (14.01% weight/volume of Na2SO4) × (9.9 mL Na2SO4 solution) / (Molar mass of Na2SO4)

Now, since the reaction ratio is 2:1, we need twice as many moles of AgNO3 to react with Na2SO4:

Moles of AgNO3 required = 2 × Moles of Na2SO4

To find the volume of AgNO3 solution required, you divide the moles of AgNO3 by the concentration of AgNO3:

Volume of AgNO3 solution = Moles of AgNO3 / (4.69% weight/volume of AgNO3)

Converting the 4.69% w/v solution to molarity and then calculating the volume gives you approximately 70.7345 mL.