J.R. S. answered • 11/08/17

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Ph.D. in Biochemistry--University Professor--Chemistry Tutor

You MUST write a balanced equation for this reaction:

2NaOH + H

_{2}SO_{4}==> Na_{2}SO_{4}+ 2H_{2}ONext, calculate moles NaOH used:

0.760 L NaOH x 0.300 moles/L = 0.228 moles (assuming you meant 0.300 M and not 0.300 m. See below)

Next, note the mole ratio in the balanced equation between NaOH and H

_{2}SO_{4}. It is 2:1Calculate moles H2SO4 present using this ratio:

0.228 moles NaOH x 1 mole H

_{2}SO_{4}/2 moles NaOH =**0.114 moles H**_{2}SO_{4}neutralized.**NOTE:**There is a significant difference between 0.3 m and 0.3 M. The former means 0.3 moles/kg of solvent and the latter means 0.3 moles/L of solution. Be careful which one you use. If the problem REALLY said 0.3 m, you would have to know the density of the NaOH to solve the problem.