Given the reaction:
BaCl2(aq) + K2CrO4(aq) BaCrO4(s) + 2KCl(aq)
Calculate the following.
(1) The grams of BaCrO4 that can be obtained from 100.0 mL of 0.373 M BaCl2.
(2) The volume of 1.00 M BaCl2 needed to react with 50.0 mL of 0.413 M K2CrO4 solution.
2007-07-20 14:40:48 · 3 answers · asked by lulu 1 in Science & Mathematics ➔ Chemistry
Let the BaCl2 solution be called S. Let BaCl2 be called B. Let BaCrO4 be called C.
Atomic wts.: Ba=137 Cr=52 O=16 BaCrO4=253
(1) 100.0mLB x 0.373molB/1000mLS x 1molC/1molB x 253gC/1molC = 9.44g BaCrO4
(2) 50.0mLK2CrO4Soln x 0.413molK2CrO4/1000mLSoln x 1molBaCl2/1molK2CrO4 x 1000mLBaCl2Soln/1.00molBaCl2 = 20.65 mL BaCl2 soln
2007-07-20 14:56:10 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋
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2007-07-20 14:42:54 · answer #2 · answered by Anonymous · 0⤊ 0⤋
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2007-07-20 14:43:10 · answer #3 · answered by koolgrl 2 · 0⤊ 0⤋