For the reaction shown above, what is the chemical formula for the limiting reac
ID: 962276 • Letter: F
Question
For the reaction shown above, what is the chemical formula for the limiting reactant? Write the balanced chemical equation for the reaction, using lowest whole-number coefficients. An aqueous solution containing 5.27 g of lead(II) nitrate is added to an aqueous solution containing 5.72 g of potassium chloride to generate solid lead(II) chloride and potassium nitrate. Write the balanced chemical equation for this reaction. Be sure to include all physical states. What is the limiting reactant? lead(I) nitrate potassium chloride The percent yield for the reaction is 91 4%, how many grams of pecipitate were recovered? How many grams of the excess reactant remain?Explanation / Answer
Q1)
the balanced equation is 2 NO + H2 ------. N2 + H2O
Q2)
Pb(NO3)2 + 2KCl ------> 2KNO3 + PbCl2(s)
5.27g/331.2 5.72/74.5 0 0 initial moles
0 0.0448 0.0318 0.0159 after reaction [ lead nitrate is the limiting reagent as it is consumed completely in the reaction ]
Theoretical yield of precipitate = 0.0159x 278.2 = 4.4266g
Experimental yild is 91.4% , weight of precipitate = 4.4266x 91.4/100 = 4.045g
The excess reagent remaining is = 0.0448 x 74.5 = 3.345 g
Q3)
Al2O3 + 6NaOH + 12HF --------> 2 Na3AlF6 + 9H2O
17.6/102 60.4/40 60.4/40 0 0 initial moles
0 0.4747 0.95 0.345 1.55 after reaction
the limiting reagent is alumina
The amount of cryolite obtained from the reaction is = 0.345x 210 = 72.45 Kg
Mass of NaOH left = 0.4747x 40 = 18.988Kg
Mass of HF left = 0.95x 20 = 19Kg
Q4)
Pb(NO3)2 + 2KCl ------> 2KNO3 + PbCl2(s)
7.19g/331.2 5.35/74.5 0 0 initial moles
0 0.02838 0.04341 0.0217 after reaction [ lead nitrate is the limiting reagent as it is consumed completely in the reaction ]
Weight of lead chloride precipitate formed = 0.0217x 278.2 x86.6/100 = 5.227g
Weight of excess reactant remaining = 0.02838x 74.5 = 2.1143g
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