If 15.50 of Pb( No3)2 are reacted with 3.81g of NaCl then NaNo3 and with PbCl are formed calculate the Mass of the excess reactants
To solve the problem, we first need to write the balanced chemical equation for the reaction between Pb(No3)2 and NaCl:
Pb(No3)2 + 2 NaCl → 2 NaNo3 + PbCl2
From the equation, we can see that 1 mole of Pb(No3)2 reacts with 2 moles of NaCl to give 2 moles of NaNo3 and 1 mole of PbCl2. Therefore, we need to calculate how many moles of Pb(No3)2 and NaCl are being used in the reaction.
The molar mass of Pb(No3)2 is:
207.2 g/mol (atomic mass of Pb = 207.2, atomic mass of N = 14.01, atomic mass of O = 16.00)
So, 15.50 g of Pb(No3)2 is equivalent to:
15.50 g / 207.2 g/mol = 0.0748 mol
The molar mass of NaCl is:
58.44 g/mol (atomic mass of Na = 22.99, atomic mass of Cl = 35.45)
So, 3.81 g of NaCl is equivalent to:
3.81 g / 58.44 g/mol = 0.0652 mol
Since the stoichiometric ratio between Pb(No3)2 and NaCl is 1:2, we can see that NaCl is the limiting reactant in the reaction. Therefore, all the Pb(No3)2 will react completely and there will be some NaCl left over.
From the equation, we can see that 2 moles of NaCl are used for every 1 mole of Pb(No3)2. So, for the 0.0748 mol of Pb(No3)2, we need:
2 x 0.0748 mol = 0.1496 mol of NaCl
The number of moles of NaCl we actually have is 0.0652 mol. So, the amount of NaCl that is used up in the reaction is:
0.0652 mol
The amount of NaCl that is left over is:
0.0652 mol - 0.1496 mol = -0.0844 mol
This negative number means that there is no excess NaCl left over - all of the NaCl has been used up in the reaction.
To determine the mass of the excess reactant, we first need to find out which reactant is the limiting reactant. The limiting reactant is the one that gets completely consumed in the reaction, thus determining the maximum amount of product that can be formed.
Let's calculate the number of moles for each reactant:
1. Moles of Pb(No3)2:
Given mass of Pb(No3)2 = 15.50 g
Molar mass of Pb(No3)2 = molar mass of Pb + 2(molar mass of NO3)
= (207.2 g/mol) + 2[(14.0 g/mol) + (16.0 g/mol) x 3]
= 207.2 g/mol + 2[(14.0 g/mol) + (48.0 g/mol)]
= 207.2 g/mol + 2[(14.0 g/mol) + (48.0 g/mol)]
= 207.2 g/mol + 2[(14.0 g/mol) + (48.0 g/mol)]
= 207.2 g/mol + 2(62.0 g/mol)
= 207.2 g/mol + 124.0 g/mol
= 331.2 g/mol
Moles of Pb(No3)2 = mass / molar mass
= 15.50 g / 331.2 g/mol
= 0.04684 mol
2. Moles of NaCl:
Given mass of NaCl = 3.81 g
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol
= 58.44 g/mol
Moles of NaCl = mass / molar mass
= 3.81 g / 58.44 g/mol
= 0.06523 mol
Now, let's determine the stoichiometric ratio between the reactants based on the balanced equation. The balanced equation is:
Pb(No3)2 + 2NaCl → PbCl2 + 2NaNo3
From the balanced equation, we can see that the ratio between Pb(No3)2 and NaCl is 1:2. Therefore, for every 1 mole of Pb(No3)2, we need 2 moles of NaCl.
Since we have 0.04684 moles of Pb(No3)2 and 0.06523 moles of NaCl, the reactant with the lower moles is the limiting reactant. In this case, Pb(No3)2 is the limiting reactant because we have fewer moles of it compared to NaCl.
Now, let's calculate the moles of PbCl2 formed using the limiting reactant:
Moles of PbCl2 = moles of limiting reactant (Pb(No3)2) x stoichiometric ratio (PbCl2 / Pb(No3)2)
= 0.04684 mol x (1 mol PbCl2 / 1 mol Pb(No3)2)
= 0.04684 mol
Since the stoichiometric ratio is 1:1 between PbCl2 and Pb(No3)2, 0.04684 moles of PbCl2 are formed.
Next, let's calculate the moles of NaNo3 formed using the limiting reactant:
Moles of NaNo3 = moles of limiting reactant (Pb(No3)2) x stoichiometric ratio (NaNo3 / Pb(No3)2)
= 0.04684 mol x (2 mol NaNo3 / 1 mol Pb(No3)2)
= 0.09368 mol
We have now determined the number of moles of each product formed. To find the mass of the excess reactant, we need to subtract the moles calculated from the initial moles of the limiting reactant.
For NaCl, the initial moles were 0.06523 mol, and we used 0.04684 mol in the reaction. Therefore, the excess moles of NaCl are:
Excess moles of NaCl = initial moles - moles used
= 0.06523 mol - 0.04684 mol
= 0.01839 mol
Finally, let's calculate the mass of the excess reactant:
Mass of excess reactant = excess moles of NaCl x molar mass of NaCl
= 0.01839 mol x 58.44 g/mol
= 1.074 g (rounded to three decimal places)
Therefore, the mass of the excess reactant (NaCl) is approximately 1.074 grams.