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
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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can you do these two questions. I am not sure If what I did was correct
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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