What mass in grams of #"HCl"# reacts completely with #"3.208 g Pb"# in the following reaction?

You start with a balanced equation. The coefficients in front of the formulas give the mole ratios of the reactants and products. You will need to convert the mass of lead (Pb) into moles of lead using the molar mass of lead. The molar mass of lead is its atomic weight on the periodic table in g/mol. Once you know how many moles of lead reacted with the HCl, you caan use the mole ratio between Pb and HCl, and the molar mass of HCl to determine the mass of HCl that would react completely with the given mas of Pb.

Balanced equation: #"Pb(s")# + #"2HCl(aq")# #rarr# #"PbCl"_2("aq")# + #"H"_2("g")#

Given/Known:
mass of Pb = 3.208 g
molar mass of Pb = 207.2 g/mol
molar mass of HCl = (1 x 1.00794) + (1 x 35.453) = 36.461g/mol
mole ratio Pb to HCl = 1 mol Pb : 2 mol HCl

Solution.

Convert mass of Pb to moles using the molar mass of Pb.

#"3.208 g Pb"# x #"1 mol Pb"/"207.2 g Pb"# = #"0.01548 mol Pb"#

Use the moles of Pb that reacted and the mole ratio of Pb and HCl to determine the number of moles of HCl that reacted. Write the mole ratio with HCl on top, which will cancel mol Pb and leave mol HCl.

#"0.01548 mol Pb"# x #"2 mol HCl"/"1 mol Pb"# = #"0.03096 mol HCl"#

Convert the moles of HCl that reacted to mass using its molar mass.

#"0.03096 mol HCl"# x #"36.461 g HCl"/"1 mol HCl"# = #"1.129 g HCl"#

Answer:
1.129 g HCl react completely with 3.208 g Pb.

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