Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfur

Question

Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to BaO2(s)+H2SO4(aq)--->BaSO4(s)+H2O2(aq) How many milliliters of 4.75 M H2SO4(aq) are needed to react completely with 44.5 g of BaO2(s)?

Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 19.01 g of PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.

Calculate the molarity of the Pb(NO3)2(aq) solution

Explanation / Answer

BaO2(s)+H2SO4(aq)--->BaSO4(s)+H2O2(aq)

no of moles of BaO2 = W/G.M.Wt   = 44.5/169.33 = 0.26 moles

From balanced 1 mole of BaO2 react with 1 mole of H2So4

                             0.26 mole of BaO2 react with 0.26 moles of H2So4

molarity of H2SO4   = no of of mole/volume in L

                  4.75      = 0.26/volume in L

volume in L   = 0.26/4.75 = 0.0547L     = 54.7ml >>>> answer

Pb(NO3)2 +2 NaCl -------> PbCl2 + 2NaNO3

no of moles of PbCl2 = W/G.M.Wt   = 19.01/278 = 0.068moles

From balanced equation 1 mole of Pb(NO3)2 gives 1 mole of PbCl2

                                          0.068 moles of Pb(NO3)2 gives 0.068 moles of PbCl2

molarity    = no of moles/volume in L

                 = 0.068/0.2 0.34M >>>> answer

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