The following questions relate to the precipitation of chloride salt (e.g. NaCl)

Question

The following questions relate to the precipitation of chloride salt (e.g. NaCl) using lead (II) nitrate. Make sure to pay attention to significant figures when providing answers.

A solution of lead (II) nitrate was added to an aqueous sample of a simple chloride salt. The resulting solid that formed, lead (II) chloride, was filtered, dried, and weighed. The mass of the lead (II) chloride was found to be 14.84 g. How many moles of lead (II) chloride precipitated from the solution?

Assuming all of the chloride ions in the original salt solution were precipitated out as lead (II) chloride, how many moles of chloride ions were in the salt sample?

What is the mass of chloride ions in the original solution based on the moles of chloride ion precipitated?

Please show me how to do this.... Step by step

Explanation / Answer

Let us write the balanced chemical equation first

NaCl (aq) + Pb(NO3)2 (aq) --> PbCl2(s) + 2NaNO3 (aq)

a) the mass of lead chloride formed = 14.84 grams

The molecular weight of lead chloride = 278.1 g/mol

So moles of lead chloride formed = Mass / Mol weight = 14.84 / 278.1 = 0.0534 moles

b) each moles of lead chloride consists of two moles of chloride (which is the total number of moles in present initially)

Moles of chloride ions = 2 X moles of PbCl2 = 2 X 0.0534 = 0.1068 moles

c) Mass of chlorde ions = Moles of chloride ions X atomic weight of chloride ions = 0.1068 X 35.5 = 3.79 grams

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