Hydrochloric acid reacts with lead(II) nitrate according to the reaction below.2HCl (aq) + Pb(NO3)2 (aq) ⟶ 2HNO3 (aq) + PbCl2(s)If 125.0 mL of a 0.250 M HCl solution is need to complete react with a sample of Pb(NO3)2, what is the mass of PbCl2 precipitate?
Question
Hydrochloric acid reacts with lead(II) nitrate according to the reaction below.2HCl (aq) + Pb(NO3)2 (aq) ⟶ 2HNO3 (aq) + PbCl2(s)If 125.0 mL of a 0.250 M HCl solution is need to complete react with a sample of Pb(NO3)2, what is the mass of PbCl2 precipitate?
Solution
To solve this problem, we need to follow these steps:
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First, we need to find the number of moles of HCl. We can do this by multiplying the volume (in liters) by the molarity.
Moles of HCl = Volume (L) x Molarity Moles of HCl = 0.125 L x 0.250 mol/L = 0.03125 mol
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From the balanced chemical equation, we can see that the ratio of HCl to PbCl2 is 2:1. Therefore, the moles of PbCl2 produced will be half the moles of HCl.
Moles of PbCl2 = 0.03125 mol / 2 = 0.015625 mol
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Finally, we can find the mass of PbCl2 by multiplying the moles by the molar mass. The molar mass of PbCl2 is approximately 278.1 g/mol.
Mass of PbCl2 = Moles x Molar mass Mass of PbCl2 = 0.015625 mol x 278.1 g/mol = 4.345 g
So, the mass of PbCl2 precipitate is approximately 4.345 g.
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