54.9 g of solid aluminum hydroxide, Al(OH)3, is neutralized with 91.9 g of sulfuric acid, H2SO4.
a. Write a balanced equation of the reaction?
b. Which one is the limiting reagent?
c. How many grams of aluminum sulfate will be produced?
d. How much of the excess reagent(in grams) is left at the end of the reaction?
2007-12-21 12:34:40 · 1 answers · asked by Charlie W 1 in Science & Mathematics ➔ Chemistry
a. 2Al(OH)3 + 3H2SO4 ===> Al2(SO4)3 + 3H2O
b. Atomic weights: Al=27 O=16 H=1 Al(OH)3=78 S=32 H2SO4=98 Al2(SO4)3=342
54.9gAl(OH)3 x 1molAl(OH)3/78gAl(OH)3 = 0.704 mole Al(OH)3
91.9gH2SO4 x 1molH2SO4/98gH2SO4 = 0.938 mole H2SO4
0.704molAl(OH)3 x 3molH2SO4/2molAl(OH)3 = 1.056 mole H2SO4
You don't have that much H2SO4, so H2SO4 is the limiting reagent.
c. 0.938molH2SO4 x 1molAl2SO4)3/3molH2SO4 x 342gAl2(SO4)3/1molAl2(SO4)3 = 107g Al2(SO4)3
2007-12-21 12:55:03 · answer #1 · answered by steve_geo1 7 · 1⤊ 0⤋