a class reacts 50.00ml samples of 0.200 mol/L potassium phosphate solution with an excess of 0.120 mol/L lead(II) nitrate solution. Which reagent is intended to be the limiting reagent, what is he minimum volume of lead II nitrate solution required. How do you convert the molar mass of g/mol in to l/mol? help please!!!
To determine which reagent is the limiting reagent, you need to compare the molar ratios of the reactants. The balanced chemical equation for the reaction between potassium phosphate (K3PO4) and lead(II) nitrate (Pb(NO3)2) is:
2K3PO4 + 3Pb(NO3)2 -> 6KNO3 + Pb3(PO4)2
From the balanced equation, you can see that for every 2 moles of K3PO4, you need 3 moles of Pb(NO3)2. Therefore, the molar ratio of K3PO4 to Pb(NO3)2 is 2:3.
Given the concentration of the potassium phosphate solution (0.200 mol/L), you can determine the number of moles of K3PO4 in 50.00 ml:
Moles of K3PO4 = concentration × volume
= 0.200 mol/L × 0.050 L
= 0.010 mol
Now, you can calculate the number of moles of Pb(NO3)2 required for complete reaction using the molar ratio:
Moles of Pb(NO3)2 needed = (2/3) × Moles of K3PO4
= (2/3) × 0.010 mol
= 0.0067 mol
To convert the molar mass from g/mol to g/L, you need to divide the molar mass by the density (g/L) of the substance. But if you meant to convert molar mass from g/mol to mol/L, you don't need to do any conversion. The molar mass of a substance is already given in g/mol, and it directly relates to the concentration in mol/L.
To find the minimum volume of lead(II) nitrate solution required, you need to consider its concentration (0.120 mol/L) and the number of moles needed (0.0067 mol):
Volume of Pb(NO3)2 solution = Moles of Pb(NO3)2 needed / concentration
= 0.0067 mol / 0.120 mol/L
= 0.056 L
= 56.0 mL
Therefore, the minimum volume of lead(II) nitrate solution required is 56.0 mL.
To determine the limiting reagent and the minimum volume of lead(II) nitrate solution required, we need to compare the stoichiometry and amount of each reactant.
1. Limiting Reagent:
First, we need to determine which reactant will be completely consumed, limiting the amount of product formed.
The balanced chemical equation for the reaction is:
K3PO4(aq) + 3Pb(NO3)2(aq) --> 2KNO3(aq) + Pb3(PO4)2(s)
From the equation, we can see that the mole ratio between K3PO4 and Pb(NO3)2 is 1:3.
Given:
Volume of K3PO4 solution = 50.00 mL = 0.05000 L
Concentration of K3PO4 solution = 0.200 mol/L
Using the formula: moles = concentration × volume
Moles of K3PO4 = 0.200 mol/L × 0.05000 L = 0.010 mol
Since the mole ratio between K3PO4 and Pb(NO3)2 is 1:3, this means that 0.010 mol of K3PO4 will react with 0.010 mol × 3 = 0.030 mol of Pb(NO3)2.
Given:
Concentration of Pb(NO3)2 solution = 0.120 mol/L
To determine the limiting reagent, we compare the actual moles available of each reactant. Since we have only 0.030 mol of Pb(NO3)2 available, and the moles required is equal to the moles of K3PO4, it means that K3PO4 is the limiting reagent.
2. Minimum Volume of Pb(NO3)2 Solution:
To find the minimum volume of Pb(NO3)2 solution required, we need to find the volume that contains the required 0.030 mol using its concentration.
Given:
Concentration of Pb(NO3)2 solution = 0.120 mol/L
Using the formula: moles = concentration × volume
0.030 mol = 0.120 mol/L × volume
volume = 0.030 mol / 0.120 mol/L
volume = 0.250 L or 250 mL
So, the minimum volume of the lead(II) nitrate solution required is 250 mL.
3. Converting Molar Mass from g/mol to L/mol:
To convert molar mass from g/mol to L/mol, we use the density of the substance.
First, we convert grams to moles using the molar mass (g/mol) and then convert moles to volume using the density (g/L) or molar volume (L/mol).
For example:
If we have the molar mass of a substance as 58.44 g/mol, and we want to convert it to L/mol, we also need the density of the substance, let's say it is 2.70 g/L.
To convert:
molar mass (g/mol) to molar volume (L/mol), we divide the molar mass by the density.
58.44 g/mol / 2.70 g/L = 21.64 L/mol
So, the molar mass of 58.44 g/mol is equivalent to 21.64 L/mol.
Remember to use the density specific to the substance you are converting.