1.How many grams of solid barium sulfate form when 19.4 mL of 0.160 M barium chloride reacts with 58.6 mL of 0.055 M sodium sulfate? Aqueous sodium chloride is the other product.
2007-09-20 06:53:46 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
Hold on i'll call Einstein and find out.
2007-09-20 06:57:46 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Atomic weights: Ba=137 S=32 O=16 Na=23 S=32 BaSO4=233
BaCl2(aq) + Na2SO4(aq) ===> BaSO4(s) + 2NaCl(aq)
19.4mLBaCl2 x 0.160molBaCl2/1000mLBaCl2 = 0.003104 moles BaCl2
58.6mLNa2SO4 x 0.055molNa2SO4/1000mLNa2SO4 = 0.003223 moles Na2SO4
The equation says that BaCl2 and Na2SO4 react 1:1 in moles. There are fewer moles BaCl2, so that will run out first. It is the limiting reagent. Going on from there:
0.003104molBaCl2 x 1molBaSO4/1molBaCl2 x 233gBaSO4/1molBaSO4 = 0.723g BaSO4
2007-09-20 14:38:24 · answer #2 · answered by steve_geo1 7 · 0⤊ 0⤋
You probably aren't getting the right answers because chemists have better things to do than answer your questions on yahoo! answers.
2007-09-20 14:08:08 · answer #3 · answered by Xindy 4 · 0⤊ 0⤋
i got about 3.48 mg... you want the solution or just want the answer?
answering questions is the whole point of yahoo answers silly!
2007-09-20 14:11:08 · answer #4 · answered by Koozie the chemist 4 · 0⤊ 0⤋