How many moles of NaCl , if mixed with excess Pb2+ ions in solution, would be needed to form 45.5 g of PbCl2 ?

The number of mole of NaCl needed to react with excess Pb²⁺ to produce 45.5 g of PbCl₂ is 0.328 mole

We'll begin by calculating the number of mole in 45.5 g of PbCl₂. This can be obtained as follow:

Mass of PbCl₂ = 45.5 g

Molar mass of PbCl₂ = 207 + (35.5×2) = 278 g/mol

Mole = mass / molar mass

Mole of PbCl₂ = 45.5 / 278

2NaCl + Pb²⁺ —> PbCl₂ + 2Na⁺

From the balanced equation above,

2 moles of NaCl reacted to produce 1 mole of PbCl₂

Therefore,

Xmol of NaCl will react to produce 0.164 mole of PbCl₂ i.e

Xmol of NaCl = 2 × 0.164

Xmol of NaCl = 0.328 mole

Thus, the number of mole of NaCl needed for the reaction is 0.328 mole

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