maximum mass of Ca(CN)2 that can be obtained from 1.56 g of HCN and 2.58 g of hoose the closest answer. sses 2 HCN(aq) + 27.03 Ca(OH)2(aq) -> Ca(CN)₂(aq) + 74.10 92.12 7HCN = 1.56g 27.03 g moi" - 0.05111 mol in Ca(OH)₂ = 2.58g 74.10 g moll 2 H₂O(1) 18.02 } This is the Limiting reagent. = 0.03482 mol I would need 0.06964 mol 8 HCN. We don't have enough
maximum mass of Ca(CN)2 that can be obtained from 1.56 g of HCN and 2.58 g of hoose the closest answer. sses 2 HCN(aq) + 27.03 Ca(OH)2(aq) -> Ca(CN)₂(aq) + 74.10 92.12 7HCN = 1.56g 27.03 g moi" - 0.05111 mol in Ca(OH)₂ = 2.58g 74.10 g moll 2 H₂O(1) 18.02 } This is the Limiting reagent. = 0.03482 mol I would need 0.06964 mol 8 HCN. We don't have enough
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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Question
How is HCN the limiting reagent if Ca(OH)2 is producing less?

Transcribed Image Text:maximum mass of Ca(CN)2 that can be obtained from 1.56 g of HCN and 2.58 g of
hoose the closest answer.
sses
2 HCN(aq) + Ca(OH)₂(aq) →→→ Ca(CN)2(aq) + 2 H₂O(1)
27.03
74.10
92.12
18.02
MHCN =
n
1.56g
27.03 g moi""
inca (CN)2
инси
m
in ca(OH)₂ = 2.589
74.10 g moil
= 0.05111 mol
ca (CN)₂ = 4 xM =
=—=— con
2
=>
= nxm = 2.66g
This is the
} Limiting reagent.
= 0.03482 mol
Lowould need 0.06964
HCN.: We don't
n
ca (CN)₂ = 0.05111m) = 0.028855mol
mol
2
mol
8
have enough
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