If 75.0 mL of an AgNO3 solution reacts with enough Cu to produce .25g Ag by single displacement, what is the molarity of the initial AgNO3 solution if Cu(NO3)2 is the other product?
Write the equation and balance it.
M Ag = moles/L. Plug and chug.
Im Lost i wrote the equation but what do i do next?
2AgNO3+Cu=2Ag+Cu(NO3)2
I gave you the equation and you don't need the equation. I just wanted to make sure you knew how to write it.
moles Ag = 0.25 g/atomic mass.
M = moles/L.
You get calculate the moles Ag deposited. Then M = moles/L.
To find the molarity of the initial AgNO3 solution, we need to use the given information and some stoichiometry calculations.
First, let's write the balanced chemical equation for the reaction:
2 AgNO3 + Cu → Cu(NO3)2 + 2 Ag
From the equation, we can see that 2 moles of AgNO3 react with 1 mole of Cu to produce 2 moles of Ag.
We are given that 0.25 g of Ag is produced, but we need to convert this mass into moles. The molar mass of Ag is approximately 107.87 g/mol, so:
moles of Ag = mass of Ag / molar mass of Ag
moles of Ag = 0.25 g / 107.87 g/mol
Now, we can use the molar ratio from the balanced equation to determine the moles of AgNO3 that reacted:
moles of AgNO3 = (moles of Ag) / (2 moles of AgNO3 / 2 moles of Ag)
moles of AgNO3 = (0.25 g / 107.87 g/mol) / (2 mol AgNO3 / 2 mol Ag)
Next, we need to convert the volume of the AgNO3 solution into liters to calculate its molarity. We are given that the volume is 75.0 mL:
volume of AgNO3 solution = 75.0 mL = 0.075 L
Now we can calculate the molarity of the initial AgNO3 solution:
molarity = moles of AgNO3 / volume of AgNO3 solution
molarity = moles of AgNO3 / 0.075 L
Finally, substitute the calculated value of moles of AgNO3 into the equation:
molarity = (0.25 g / 107.87 g/mol) / 0.075 L
Calculating this expression will give you the molarity of the initial AgNO3 solution.