Question 6: When you and your partner are finished with the above work, solve th

Question

Question 6: When you and your partner are finished with the above work, solve the following problems What is the minimum amount of 6.0 M H2SO, necessary to produce 25.0 g of hydrogen according to the unbalanced chemical equation? Balance it and add the physical states before doing the calculations. A 55.0-mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead(lI) acetate solution and this precipitation reaction occurs: 2 KC HO(ag) + PbSO(s) The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Explanation / Answer

2Al(s) + H2SO4(aq) ----> Al2(SO4)3(aq) + H2(g)

2 mol Al = 1 MOL h2so4 = 1 mol H2

NO of mol of H2 = 25/2 = 12.5 mol

NO of mol of H2SO4 required = 12.5 mol

Amount of 6.M h2so4 required = n/M = 12.5/6 = 2.083 L

2) FROM EQUATION , 1 mol K2SO4 = 1 mol Pb(C2H3O2)2 = 1 mol PbSO4

   no of mol of K2SO4 = 55*0.102/1000 = 0.00561 mol

   no of mol of Pb(C2H3O2)2 = 35*0.114/1000 = 0.004 mol

   Limiting reactant = Pb(C2H3O2)2

no of mol of PbSO4 formed = 0.004 mol

Amount of PbSO4 formed = 0.004*303.26

    Theoretical yield of PbSO4 = 1.213 g

        practical yield of PbSO4 = 1.01 g

percent yield = practical yield / Theoretical yield*100

          = 1.01/1.213*100

   = 83.265%

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