-What volume of .500 M Pb(NO3)2 is required to react completely with 18.25 L of .625 M Na2SO4?
Pb(NO3)2 + Na2SO4 ==> PbSO4 + 2NaNO3
mols Na2SO4 = M x L = ?
mols Pb(NO3)2 = 1/2 mol Na2SO4
M Pb(NO3)2 = mols Pb(NO3)2/L Pb(NO3)2. You have M and mols, solve for L.
thank you..
To determine the volume of Pb(NO3)2 required to react completely with Na2SO4, we need to use the concept of stoichiometry.
The balanced chemical equation for the reaction between Pb(NO3)2 and Na2SO4 is:
Pb(NO3)2 + Na2SO4 -> PbSO4 + 2NaNO3
From the equation, we can see that 1 mole of Pb(NO3)2 reacts with 1 mole of Na2SO4.
First, let's calculate the number of moles of Na2SO4 present in 18.25 L of a 0.625 M solution.
Molarity (M) is defined as moles of solute per liter of solution.
Given:
Volume of Na2SO4 solution = 18.25 L
Molarity of Na2SO4 solution = 0.625 M
Number of moles of Na2SO4 = Molarity x Volume
= 0.625 mol/L x 18.25 L
= 11.40625 mol
According to the balanced equation, 1 mole of Pb(NO3)2 reacts with 1 mole of Na2SO4. Therefore, the number of moles of Pb(NO3)2 required for complete reaction is also 11.40625 mol.
Now, we need to convert moles of Pb(NO3)2 to volume using its molarity.
Given:
Molarity of Pb(NO3)2 = 0.500 M
Volume of Pb(NO3)2 = Moles / Molarity
= 11.40625 mol / 0.500 mol/L
= 22.8125 L
Therefore, the volume of 0.500 M Pb(NO3)2 required to react completely with 18.25 L of 0.625 M Na2SO4 is 22.8125 L.