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 of PbCl₂ =?
Mole = mass / molar mass
Mole of PbCl₂ = 45.5 / 278
Mole of PbCl₂ = 0.164 mole
- Finally, we shall determine the number of mole of NaCl needed to produce 0.164 mole (i.e 45.5 g) of PbCl₂. This can be obtained as follow:
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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