Question

If

75.0 mL of an AgNO3 solution reacts with enough Cu to produce a 0.250 g Ag by
ngle displacement, what is the molarity of the initial AgNO, solution if Cu(NO3)2 is the
ther product?

Answer 1

The balanced chemical equation for the reaction is:

2AgNO3 + Cu -> Cu(NO3)2 + 2Ag

From the equation, we can see that 2 moles of AgNO3 react with 1 mole of Cu to produce 2 moles of Ag.

First, we need to calculate the number of moles of Ag produced:

0.250 g Ag * (1 mol Ag / 107.87 g Ag) = 0.00232 mol Ag

Since 2 moles of AgNO3 produce 2 moles of Ag, we know that 0.00232 mol Ag was produced from 0.00232 mol AgNO3.

Next, we can calculate the molarity of the initial AgNO3 solution:

Molarity = moles of solute / liters of solution

We know that 75.0 mL of the AgNO3 solution was used in the reaction, which is 0.0750 L. We also know that 0.00232 mol of AgNO3 was used in the reaction.

Molarity = 0.00232 mol / 0.0750 L = 0.031 M

Therefore, the molarity of the initial AgNO3 solution is 0.031 M.