What mass of NaOH is required to react exactly with 25.0 mL of 1.2 M H2SO4?
A. 1.2 g
B. 1.8 g
C. 2.4 g
D. 3.5 g
E. none of these
2007-02-07 13:25:21 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
H2SO4 + 2NaOH ===> Na2SO4 + H2O
Let sulfuric acid solution be called S. Let H2SO4 be called H. Let NaOH be called N. Molecular weight of NaOH = 40
25.0mLS x 1.2molH/1000mLS x 2molN/1molH x 40gN/1molN = (25.0)(1.2)(2)(40)/(1000) = 2.4g NaOH
The 25.0mL is given. The first factor comes from the molarity. The mL S cancel, leaving moles of H2SO4. The second factor comes from the balanced equation. The moles H2SO4 cancel, leaving moles NaOH. The third factor comes from the molecular weight of NaOH. The moles NaOH cancel, leaving grams NaOH.
2007-02-07 13:43:03 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋