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When 4.56 L of a 0.0123 M Al(NO3)3 solution reacts with 220. mL of a 0.321 M Na2SO3 solution, 5.40 g of a precipitate is obtained. solid? actual 1. Write a formula unit equation for this precipitation reaction. * + (۸) 43 (-1)3 2Al(NO3)₂ 3(aq) 42 + + 3 Na₂ 103 (a s→ (og) (aq) + Al (103) (1) 6 Na NO3 (ag) 2. Calculate the percent yield for this reaction.

When 4.56 L of a 0.0123 M Al(NO3)3 solution reacts with 220. mL of a 0.321 M Na2SO3 solution, 5.40 g of a precipitate is obtained. solid? actual 1. Write a formula unit equation for this precipitation reaction. * + (۸) 43 (-1)3 2Al(NO3)₂ 3(aq) 42 + + 3 Na₂ 103 (a s→ (og) (aq) + Al (103) (1) 6 Na NO3 (ag) 2. Calculate the percent yield for this reaction.

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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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When 4.56 L of a 0.0123 M Al(NO3)3 solution reacts with 220. mL of a 0.321 M Na2SO3 solution, 5.40 g of a precipitate is obtained. solid? actual 1. Write a formula unit equation for this precipitation reaction. L→ (1) +3 (-1)3 2 Al (NO3)2 (aq) +2 -2 + 3 Na₂ 103 (₂) s→ (ag) 220 ml Na₂SO3 x 2. Calculate the percent yield for this reaction. 4.56L Al(103) x 0.0123 mol Al (NO3)₂ 3 1 L Al (NO3)3 12 1000 ml 0.0235 mol A₂ (80₂) z x X % yield = x 246 g Al₂(103)3 1 mol Al₂ (103) ₂ +3 -2 +1 NO₂ Al (103) (4) + 6 Na JrOs (ag) 0.321 mol Na₂SO3 12 Naz Soz 5.40g Al₂ (S03)3 5.18g Al2 (103)3 1 mol Al(103)3 = 0.0280 mol Al (103) 3 2 mol Al (Jos) X = 5.18g Al₂ (103)3 X100 = 93.4% Limited Reagen 1 mol Al₂(SO3)3 = 0.0235 mol A1₂ (603) 3 mol Naz Soz
Transcribed Image Text:When 4.56 L of a 0.0123 M Al(NO3)3 solution reacts with 220. mL of a 0.321 M Na2SO3 solution, 5.40 g of a precipitate is obtained. solid? actual 1. Write a formula unit equation for this precipitation reaction. L→ (1) +3 (-1)3 2 Al (NO3)2 (aq) +2 -2 + 3 Na₂ 103 (₂) s→ (ag) 220 ml Na₂SO3 x 2. Calculate the percent yield for this reaction. 4.56L Al(103) x 0.0123 mol Al (NO3)₂ 3 1 L Al (NO3)3 12 1000 ml 0.0235 mol A₂ (80₂) z x X % yield = x 246 g Al₂(103)3 1 mol Al₂ (103) ₂ +3 -2 +1 NO₂ Al (103) (4) + 6 Na JrOs (ag) 0.321 mol Na₂SO3 12 Naz Soz 5.40g Al₂ (S03)3 5.18g Al2 (103)3 1 mol Al(103)3 = 0.0280 mol Al (103) 3 2 mol Al (Jos) X = 5.18g Al₂ (103)3 X100 = 93.4% Limited Reagen 1 mol Al₂(SO3)3 = 0.0235 mol A1₂ (603) 3 mol Naz Soz
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