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.
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
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1 answer