When 27.5 mL of 0.615 M H2SO4 is added to 27.5 mL of 1.23 M KOH in a coffee-cup

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Explanation / Answer

First write down the balanced chemical reaction: H2SO4 + 2 KOH -> K2SO4 + H2O Then calculate how many moles of each reactant you have: moles H2SO4 = (0.0275L)*(0.615M) = 0.0169125 moles H2SO4 moles KOH = (0.0275L)*(1.23M) = 0.033825 moles KOH From this information determine which reactant, if any, runs out before the other: Since it requires 2 moles of KOH for every mole of H2SO4 you see that both will be used up in the reaction. Next, calculate the energy released from: E=mc(T2-T2) = (27.5ml + 27.5ml)*(1g/ml)*(1cal/gC)*(30.17C-23.50C)

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