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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
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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How is HCN the limiting reagent if Ca(OH)2 is producing less?

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
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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