Question

0.8 pts Question 5 Complete neutralization of 25.00 mL of 0.15 15 M H2SO4 solution requires moles of KOH and forms grams H2O. (Hint: write the balanced equation first) (Report your answer with 4 significant figures. Do NOT express your answer in scientific notation) (Molar Mass H2O = 18.016 g/mol)

Answer 1

H2SO4 + 2KOH --> K2SO4 + 2 H2O

given
volume mof H2SO4 = 25 ml = 0.025 L
molarity = 0.1515 M

So number of mols of H2SO4 present = molarity*volume = 0.1515 M*0.025 L =
= 0.0037875 mols

from balanced equation it is clear that 1 mol H2SO4 neeed 2 mol KOH to neutrlize

So 0.0037875 mol H2SO4 require 2*0.0037875 mol KOH
which is =0.007575 mols

mols of KOH needed = 0.007575 mols
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Also 1 mole H2SO4 produce 2 mole H2O
Thus moles of H2O produced =0.007575 mols
Mass of H2O formed = moles of H2O* molar mass of H2O
=0.007575 mols *18.016 g/mol =0.1364712 g
in 4 sig.figures,
Mass of H2O formed =0.1365 g
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:) Thanks!