if you begin with 175 g of Pb(NO3)2 and 80 g of NaCl which is your limiting reac

Question

if you begin with 175 g of Pb(NO3)2 and 80 g of NaCl which is your limiting reactant



How many grams of PbCl2 can you make?


What is the percent yield if you produce 127g of PbCl2 in your experiment if you begin with 175 g of Pb(NO3)2 and 80 g of NaCl which is your limiting reactant



How many grams of PbCl2 can you make?


What is the percent yield if you produce 127g of PbCl2 in your experiment if you begin with 175 g of Pb(NO3)2 and 80 g of NaCl which is your limiting reactant



How many grams of PbCl2 can you make?


What is the percent yield if you produce 127g of PbCl2 in your experiment

Explanation / Answer

Solution :- Balanced reaction equation

Pb(NO3)2 (aq) + 2 NaCl(aq) ------------ > PbCl2(s) + 2NaNO3(aq0

Lets calculate the mass of the PbCl2 that can be produced from each reactant

Calculating from the Pb(NO3)2

(175 g Pb(NO3)2 * 1 mol / 331.2 g) *(1 mol PbCl2/1 mol Pb(NO3)2)*(278.1 g / 1 mol PbCl2) = 146.5 g PbCl2

Now lets calculate from the NaCl

(80 g NaCl* 1 mol / 58.443 g)*(1mol PbCl2 / 2 mol NaCl)*(278.1 g / 1 mol PbCl2) =190 g PbCl2

Here Pb(NO3)2 gives the less mass of the PbCl2 therefore Pb(NO3)2 is the limiting reactant

Therefore the theoretical yield (mass can be produced) of the PbCl2 = 146.5 g

Now lets calculate the percent yield of the PbCl2

Percent % yield = (actual yield / theoretical yield )*100%

                            = (127 g / 146.5g)*100%

                          = 86.69 %

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