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How many grams of lead 2 chloride can be produced from 14.6 liters of chlorine gas at STP and 285 milliliters of 0.250 molarity lead 2 nitrate?
Jesusibarra902
The formula unit for lead (II) chloride is Pb(NO3)2. This formula shows that each formula unit contain one lead (II) ion. By the definition of molarity, a liter of a 0.250 molar solution of lead (II) nitrate therefore contains 0.25 gram formula units of lead (II) ions per liter of solution, so that 250 milliliters of such a solution will contain 250/1000 of this amount, or 0.0625 gram formula units.
Assuming that chlorine is an ideal gas, 22.4 liters of it at STP contain one mole. Since gases are homogeneous, 14.6 liters of it will contain 14.6/22.4 or 0.652 moles. The formula of gaseous chlorine at STP is Cl2. Therefore, 0.652 moles of it contain twice this number of chlorine atoms, or 1.305 "moles" of such atoms.
The formula unit for lead (II) chloride is PbCl2. Therefore, each gram formula unit of lead (II) requires two gram formula units of chloride ions.
Comparison shows that there is a large excess of chlorine over that required to form lead (II) chloride from the 0.0625 gram formula units of lead (II) contained in the specified amount of solution. Lead (II) is therefore the limiting reagent in this combination, and 0.0625 gram formula units of PbCl2 can be produced.
The gram Atomic Mass of lead is 207.2 and that of chlorine is 35.453. Therefore, the mass of lead (II) chloride that can be produced is 0.0625[207.2 + 2(35.453)] or 17.4 grams, to the justified number of significant digits.
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