Asked by Annie
What is the concentration of acid or base remaining after reaction if 21 ml of 0.54 M sulfuric acid is mixed with 39 milliliters of 0.12 M sodium hydroxide?
Answers
Answered by
DrBob222
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
millimols H2SO4 = 21 x 0.54 = approx 11 but you need a better answer than that.
mmols NaOH = 39 x 0.12 = estimated 5. Which is the limiting reagent. 5 mmols NaOH will require 2.5 mmols H2SO4. You have that much; therefore, NaOH is the limiting reagent. Subtract 11-2.5 = estimated 8.5 mmols H2SO4 left after the reaction. Again this is an estimate. (H2SO4) = millimols/total mL
Note total mL = 39 + 21 = ?
millimols H2SO4 = 21 x 0.54 = approx 11 but you need a better answer than that.
mmols NaOH = 39 x 0.12 = estimated 5. Which is the limiting reagent. 5 mmols NaOH will require 2.5 mmols H2SO4. You have that much; therefore, NaOH is the limiting reagent. Subtract 11-2.5 = estimated 8.5 mmols H2SO4 left after the reaction. Again this is an estimate. (H2SO4) = millimols/total mL
Note total mL = 39 + 21 = ?
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