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4. When 53 mL of 0.75 M cobalt (III) nitrate are added to a sodium sulfate solution, how many grams of cobalt (III) sulfate can be precipitated? 53 mL XL X 1000ML 2Co(NO3)3 + 3Na2SO4 → Co₂(SO4)3 + 6NaNO3 20ONE imor XX 2 SO maro NO 2103 75 mal IL 1162935 7 - و426 TM ol SR grams of colbat (III) sulfate

4. When 53 mL of 0.75 M cobalt (III) nitrate are added to a sodium sulfate solution, how many grams of cobalt (III) sulfate can be precipitated? 53 mL XL X 1000ML 2Co(NO3)3 + 3Na2SO4 → Co₂(SO4)3 + 6NaNO3 20ONE imor XX 2 SO maro NO 2103 75 mal IL 1162935 7 - و426 TM ol SR grams of colbat (III) sulfate

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Do I multiply the numbers in front of the solutions when doing the Stoichiometry process. For instance, in number 4 it’s: 1 mol 3Na2SO4, do you multiply the 3 when doing the equation?
2. 3. 450M 175 mLx 4. 5. ML K 23.5 mL x R 1000m2 +96€ L 1000ML 45.3 mx L X x 100034 ML X 1000 ML 366.7 How many grams of lead (II) nitrate are needed to fully react 23.5 mL of 0.55 M sodium chloride in the precipitation of lead (II) chloride? 1.50 서이 IL * 2(NH₂), • som 1 X Pb(NO3)2+2NaCl → PbCl₂ + 2NaNO3 -55 mol 2 Nach 53 mL x L x 575 mai 200 NE 13. X 1000ML IL SOMSON IL Na₂SO4 + 2H₂O 207+28+१७ 4.3 grams. How many mL of 0.40 M barium chloride are needed to react 175 mL of 1.50 M ammonium phosphate in the precipitation of barium phosphate? 3BaCl2 + 2(NH4)3PO4 → Ba3(PO4)2 + 6NH4Cl 3Bacl2 Imol 2(UH4) 3PO4 moi 65 mol HCI X x imol x ma 2.Na DH imui Naza 42504 I MOI No 2003 IL . 40 M 2Co(NO3)3 + 3Na2SO4 → Co₂(SO4)3 + 6NaNO3 NO SON X I mol Al(OH)3 When 53 mL of 0.75 M cobalt (III) nitrate are added to a sodium sulfate solution, how many grams of cobalt (III) sulfate can be precipitated? 3 #CI 7 3 Bac 2 X Pb(NO x X IL I mol 20 - 751 TM Ol 1161935 179 How many grams of aluminum hydroxide will be neutralized by 45.3 mL of 0.55 MHCI? 3HCP+ Al(OH)3 → AlCl3 + 3H₂O 1000 ML - x x 4269= 3M 50 1000 m 3319 grams of colbat (III) sulfate
Transcribed Image Text:2. 3. 450M 175 mLx 4. 5. ML K 23.5 mL x R 1000m2 +96€ L 1000ML 45.3 mx L X x 100034 ML X 1000 ML 366.7 How many grams of lead (II) nitrate are needed to fully react 23.5 mL of 0.55 M sodium chloride in the precipitation of lead (II) chloride? 1.50 서이 IL * 2(NH₂), • som 1 X Pb(NO3)2+2NaCl → PbCl₂ + 2NaNO3 -55 mol 2 Nach 53 mL x L x 575 mai 200 NE 13. X 1000ML IL SOMSON IL Na₂SO4 + 2H₂O 207+28+१७ 4.3 grams. How many mL of 0.40 M barium chloride are needed to react 175 mL of 1.50 M ammonium phosphate in the precipitation of barium phosphate? 3BaCl2 + 2(NH4)3PO4 → Ba3(PO4)2 + 6NH4Cl 3Bacl2 Imol 2(UH4) 3PO4 moi 65 mol HCI X x imol x ma 2.Na DH imui Naza 42504 I MOI No 2003 IL . 40 M 2Co(NO3)3 + 3Na2SO4 → Co₂(SO4)3 + 6NaNO3 NO SON X I mol Al(OH)3 When 53 mL of 0.75 M cobalt (III) nitrate are added to a sodium sulfate solution, how many grams of cobalt (III) sulfate can be precipitated? 3 #CI 7 3 Bac 2 X Pb(NO x X IL I mol 20 - 751 TM Ol 1161935 179 How many grams of aluminum hydroxide will be neutralized by 45.3 mL of 0.55 MHCI? 3HCP+ Al(OH)3 → AlCl3 + 3H₂O 1000 ML - x x 4269= 3M 50 1000 m 3319 grams of colbat (III) sulfate
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