If 15.50g of Pb(No3)2 are reacted with 3.81g of NaCl then NaNo3 and with PbCl are formed what is the mass of PbCl
To solve this problem, we need to use stoichiometry and the concept of limiting reactant.
1. Write the balanced chemical equation for the reaction:
Pb(No3)2 + 2NaCl -> 2NaNo3 + PbCl2
2. Calculate the number of moles of each reactant:
- Moles of Pb(No3)2 = 15.50g / 331.2 g/mol = 0.0468 mol
- Moles of NaCl = 3.81g / 58.44 g/mol = 0.0652 mol
3. Determine the limiting reactant:
We can see from the balanced equation that the stoichiometric ratio of Pb(No3)2 to NaCl is 1:2. Therefore, for every 1 mole of Pb(No3)2, we need 2 moles of NaCl to completely react. In this case, we have more moles of NaCl than what is required according to the ratio, so Pb(No3)2 is the limiting reactant.
4. Calculate the number of moles of the products:
- Moles of NaNo3 formed = 2 x 0.0468 mol = 0.0936 mol
- Moles of PbCl2 formed = 0.0468 mol
5. Calculate the mass of PbCl2 formed:
Mass of PbCl2 = moles x molar mass = 0.0468 mol x 278.1 g/mol = 13.0 g
Therefore, the mass of PbCl2 formed is 13.0 g.