lead (II) chloride reacts with water to form lead (II) oxide and hydrchloric acid . If the enthalpies of formation of the reactants and products are:lead(II)chloride=-359.4 kj/mole, lead(II)oxide=-217.3 kj/mole, water=-285.8 kj/mole, and hydrochloric acid=-323.9 kj/mole, what is the heat or enthalpy of the reaction?
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moles of lead nitrate = 0.858*0.0456 = 0.0391248
Moles of NaCl = 0.919*0.0957 = 0.0879483
1 mole of lead nitrate react with 2 moles of NaCl
so here limiting reagent is lead nitrate.
So 0.0391248 moles of lead nitrate produce 912 J
So 1 mole lead nitrate would produce = 23.31 kJ
So (delta)H° for this reaction is -23.31 kJ
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