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Question

how many grams of PbCl2 are produced when 1.4 moles of NaCl and 3.3 moles of Pb(NO3)2 are used?

Molar Mass- PbCl2=278.106

NaCl=58.443

Pb (NO3)2=331.208

Answer 1

First, write a balanced equation, and then determine which reactant is limiting.

Balanced equation:

Pb(NO₃)₂ + 2 NaCl ==> PbCl₂ + 2 NaNO₃

Determination of limiting reactant:

moles of Pb(NO₃)₂ present = 3.3 moles

moles NaCl present = 1.4 moles

Limiting reactant = NaCl because it takes 2 NaCl for each 1 Pb(NO₃)₂ according to the balanced equation.

Next look at the mole relationship between NaCl and PbCl₂.  It is 2 NaCl = 1 PbCl₂, so the maximum moles PbCl₂ that can be formed = 0.7.  That is:  1.4 moles NaCl x 1 mole PbCl₂/2 moles NaCl = 0.7 moles PbCl₂.

Grams of PbCl2 that are formed = 0.7 moles PbCl2 x 278.106 g/mole = 194.67 grams 

If using significant figures, the answer would round to 190 grams