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How many moles of NaCl, if mixed with excess Pb2+ ions in solution, would be needed to form 10.9 g of PbCl2 ?

User Rob Cannon
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1 Answer

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1) First, let's write the chemical equation:

2 NaCl + Pb²⁺ ---> PbCl₂ + 2 Na⁺

2) Let's calculate the number of moles in 10.9 g of PbCl₂. You can obtain using the following equation:

Moles = mass/molar mass

Mass of PbCl₂ = 10.9

Molar mass of PbCl₂ = 207 + (35.5x2) = 278 g/mol

Moles of PbCl₂ = 10.9/278

Moles of PbCl₂ = 0.039 moles

3) How many moles of NaCl are needed to form 0.04 moles of PbCl₂? To find out, we need to use the proportion of the 1) reaction.

2 moles of NaCl ---- 1 mol of PbCl₂

x moles of NaCl ---- 0.039 mol of PbCl₂

1x = 0.078

x = 0.078 moles

Answer: The number of moles of NaCl is 0.078 moles.

User Edan
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