If 15.50g of Pb(No3)2 are reacted with 3.81 of NaCl then NaNo3 and with PbCl are formed which of the reactants is in excess
To determine which reactant is in excess, we need to calculate how much of each reactant is required based on the balanced chemical equation.
The balanced chemical equation for the reaction is:
Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2
From the equation, we see that 1 mole of Pb(NO3)2 reacts with 2 moles of NaCl. So we need to calculate the number of moles of each reactant:
Number of moles of Pb(NO3)2 = 15.50 g / molar mass of Pb(NO3)2
= 15.50 g / (207.2 g/mol + 3(16.00 g/mol))
= 0.0468 mol
Number of moles of NaCl = 3.81 g / molar mass of NaCl
= 3.81 g / 58.44 g/mol
= 0.0652 mol
According to the balanced chemical equation, 1 mole of Pb(NO3)2 reacts with 2 moles of NaCl. So if we take 0.0468 moles of Pb(NO3)2, we need 2 x 0.0468 = 0.0936 moles of NaCl for complete reaction. As we only have 0.0652 moles of NaCl, it is the limiting reactant.
Therefore, NaCl is the limiting reactant and Pb(NO3)2 is in excess.